file://c:\users\buba\kaz\ouba\m_6_a_.html Page of Exercise A, Question A bird flies 5 km due north and then 7 km due east. How far is the bird from its original position, and in what direction? d = \ 5 + 7 = \ 5 + 49 = \ 74 8.60 km θ = tan 7 = tan.4 = 54.46. 5 The bird is 8.60 km (3 s.f.) from the starting point on a bearing of 054 (nearest degree).
Page of Exercise A, Question A girl cycles 4 km due west then 6 km due north. Calculate the total distance she has cycled and her displacement from her starting point. Distance cycled = 4 km + 6 km = 0 km d = \ 4 + 6 = \ 6 + 36 = \ 5 7. θ = tan 6 = tan.5 = 56.3 4 bearing = 70 + 56 = 36 The displacement is 7. km (3 s.f.) on a bearing of 36 (nearest degree). file://c:\users\buba\kaz\ouba\m_6_a_.html
file://c:\users\buba\kaz\ouba\m_6_a_3.html Page of Exercise A, Question 3 A man walks 3 km due east and then 5 km northeast. Find his distance and bearing from his original position. d = 3 + 5 3 5 cos 35 = 55. d = 7.43 km (3 s.f.) sin θ 5 = sin θ = sin 35 d 5 sin 35 d = 0.476,θ = 8.4 (3 s.f.) bearing is 90 8 = 06 (nearest degree)
file://c:\users\buba\kaz\ouba\m_6_a_4.html Page of Exercise A, Question 4 In an orienteering exercise, a team hike 8 km from the starting point, S, on a bearing of 300 then 6 km on a bearing 040 to the finishing point, F. Find the magnitude and direction of the displacement from S to F. d = 6 + 8 6 8 d = 9.3 km (3 s.f.) sin θ 6 = sin θ = sin 80 d 6 sin 80 d cos 80 = 83.3 = 0.647,θ = 40.3 (3 s.f.) bearing is 300 + 40 = 340 (nearest degree) the vector SF is 9.3 km (3 s.f.) on a bearing of 340 (nearest degree)
file://c:\users\buba\kaz\ouba\m_6_a_5.html Page of Exercise A, Question 5 A boat travels 0 km on a bearing of 060, followed by 5 km on a bearing of 0. What course should it take to return to its starting point by the shortest route? d = 0 + 5 0 5 cos 30 = 00.67 d = 3.8 km (3 s.f.) sin θ 0 = sin θ = sin 30 d 0 sin 30 d = 0.489,θ = 8.8 (3 s.f.) bearing of the start = 360 9 70 = 6 (nearest degree) the return course is 3.8 km (3 s.f.) on a bearing of 6 (nearest degree)
file://c:\users\buba\kaz\ouba\m_6_a_6.html Page of Exercise A, Question 6 An aeroplane flies from airport A to airport B 80 km away on a bearing of 070. From B the aeroplane flies to airport C, 60 km from B. Airport C is 90 km from A. Find the two possible directions for the course set by the aeroplane on the second stage of its journey. 6 + 8 9 cos θ = = 0.979 6 8 bearing θ = 78.6 (3 s.f.) of C from B is 80 + 70 θ=7.4 ( d.p.) or 80 + 70 +θ=33.6 ( d.p.)
file://c:\users\buba\kaz\ouba\m_6_a_7.html Page of Exercise A, Question 7 In a regatta, a yacht starts at point O, sails km due east to A, 3 km due south from A to B, and then 4 km on a bearing of 80 from B to C. Find the displacement vector of C from O. Starting with the displacement OB, distance OB = \ + 3 = \ 3 θ = tan 3 = 33.7 (3 s.f.) Now, looking at triangle OBC, α = 80 33.7 = 46.3 (3 s.f.) d = 4 + 3 4 \ 3 d = 3.0 km (3 s.f.) sin β 4 = sin 46.3 d cos 46.3 = 9.07 4 sin 46.3 sin β = = 0.960,β = 73.59 d bearing = 90 + ( 90 θ) +β=0 (nearest degree) vector ie OC is 3.0 km (3 s.f.) on a bearing of 0 (nearest degree)
file://c:\users\buba\kaz\ouba\m_6_b_.html Page of Exercise B, Question ACGI is a square, B is the mid-point of AC, F is the mid-point of CG, H is the mid-point of GI, and D is the mid-point of AI. b and d are represented in magnitude and direction by AB and AD respectively. Find, in terms of b and d, the vectors represented in magnitude and direction by a AC, b BE, c HG, d DF, e AE, f DH, g HB, h FE, i AH, j BI, k EI, l FB. In this exercise there will usually be several correct routes to the answers because the addition law for vectors allows several options for equivalent vectors. You might reach the correct answers by a different routes to those used in these solutions. a AC = AB = b b BE = AD (parallel and equal in length) = d c HG = BC (parallel and equal in length) = AB (B is midpoint of AC) = b d DF = AC (parallel and equal in length) = b e AE = AD + DE ( triangle law of addition ) = AD + AB ( DE and AB parallel and equal in length ) = d + b f DH = DI + IH ( triangle law of addition ) = AD + AB ( AD = DI because D is the mid F point of AI, and AB is parallel and equal to IH ) = d + b
file://c:\users\buba\kaz\ouba\m_6_b_.html Page of g HB = BH ( same length, opposite direction ) = AI ( parallel and equal in length ) = d h FE = EF ( same length, opposite direction ) = HG ( parallel and equal in length ) = b ( from part c ) i AH = AI + IH (triangle law of addition) = d + b j BI = BA + AI = AB + AI = b + d k EI = EB + BA + AI = BE AB + AI = d b + d = b + d l FB = FD + DA + AB = DF AD + AB = b d + b = b d
file://c:\users\buba\kaz\ouba\m_6_b_.html Page of Exercise B, Question OACB is a parallelogram. M, Q, N and P are the mid-points of OA, AC, BC and OB respectively. p and m are equal to OP and OM respectively. Express in terms of p and m a OA b OB c BN d DQ e OD f MQ g OQ h AD i CD j AP k BM l NO. a OA = OM (M is the mid F point of OA) = m b OB = OP (P is the mid F point of OB) = p c BN = BC = OA (opposite sides parallel and equal) = m d DQ = PD ( MN and PQ bisect each other ) = OM ( line segments parallel and equal in length ) = m e OD = OP + PD ( addition of vectors ) = OP + OM ( PD and OM are parallel and equal in length ) = p + m f MQ = MO + OP + PQ ( vector addition ) = OM + OP + OA ( PQ and OA are parallel and equal in length ) = m + p + m = p + m g OQ = OP + PQ = p + m h AD = AO + OD ( vector addition ) = OA + OD = m + ( p + m ) = p m i CD = CN + ND ( vector addition ) = MO + PO ( line segments parallel and equal in length )
file://c:\users\buba\kaz\ouba\m_6_b_.html Page of = OM + OP = m p j AP = AO + OP (vector addition) = OA + OP = m + p k BM = BO + OM (vector addition) = OB + OM = p + m l NO = NB + BO ( vector addition ) = MO + BO ( MO and NB are parallel and equal in length ) = OM + OB = m p
file://c:\users\buba\kaz\ouba\m_6_b_3.html Page of Exercise B, Question 3 OAB is a triangle. P, Q and R are the mid-points of OA, AB and OB respectively. OP and OR are equal to p and r respectively. Find, in terms of p and r a OA b OB c AB d AQ e OQ f PQ g QR h BP. Use parts b and f to prove that triangle PAQ is similar to triangle OAB. a OA = OP (P is the mid F point of OA) = p b OB = OR (R is the mid F point of OB) = r c AB = AO + OB (addition of vectors) = OA + OB = p + r d AQ = AB (Q is the mid F point of AB) = p + r = p + r e OQ = OA + AQ (addition of vectors) = p + ( p + r ) = p + r f PQ = PO + OQ (addition of vectors) = OP + OQ = p + ( p + r ) = r g QR = QO + OR (addition of vectors) = OQ + OR = ( p + r ) + r = p h BP = BO + OP (addition of vectors) = OB + OP = r + p From b OB = r, and from f PQ = r, OB and PQ are parallel AOB = APQ and ABO = AQP (corresponding angles, parallel lines)
file://c:\users\buba\kaz\ouba\m_6_b_3.html Page of Angle A is common to both triangles triangles PAQ and OAB are similar (three equal angles)
file://c:\users\buba\kaz\ouba\m_6_b_4.html Page of Exercise B, Question 4 OAB is a triangle. OA = a and OB = b. The point M divides OA in the ratio :. MN is parallel to OB. Express the vector ON in terms of a and b. M divides OA in the ratio : OM = 3 a Using vector addition, ON = OA + AN = OA +λab (N lies on AB, so AN =λab ) = a +λ( a + b ) and ON = OM + MN = OM +µob (MN is parallel to OB) = 3 a +µb a +λ a + b = a +µb 3 (by comparing coefficients of a and b), λ= and λ =µ 3 so λ =µ= and ON = a + b 3 3 3
file://c:\users\buba\kaz\ouba\m_6_b_5.html Page of Exercise B, Question 5 OABC is a square. M is the mid-point of OA, and Q divides BC in the ratio :3. AP and MQ meet at P. If OA = a and OC = c, express OP in terms of a and c. M is the mid F point of OA, so OM = OA = a Using vector addition, MQ and AC = MA + AB + BQ = MA + AB + BC = a + c a = a + c = AO + OC = a + c 4 4 4 P lies on both AC and MQ, so OP = OM +λmq = a +λ a + c 4 and OP = OA +µac = a +µ( a + c ) a +λ a + c = a +µ a + c 4 by comparing coefficients of a and c, we get +λ = µ and λ =µ 4 5 λ =,λ =µ= and OP = a + b 4 5 3 5 5
file://c:\users\buba\kaz\ouba\m_6_b_5.html Page of
file://c:\users\buba\kaz\ouba\m_6_c_.html Page of Exercise C, Question Express the vectors v, v, v 3, v 4, v 5 and v 6 using the i, j notation. v = 4i, v = 5i + j, v 3 = 3i + j, v 4 = i + 3j, v 5 = i j, v 6 = 3j. Note that some people prefer to describe vectors as column vectors. In this case, the answers would be 4 0 5 3 v =, v =, v 3 =, v 4 =, v 5 =, v 6 =. 3 0 3
file://c:\users\buba\kaz\ouba\m_6_d_.html Page of Exercise D, Question Given that a = i + 3j and b = 4i j, find these terms of i and j. a a + b b 3a + b c a b d b + a e 3a b f b 3a g 4b a h a 3b a a + b = ( i + 3j ) + ( 4i j ) = ( + 4 ) i + ( 3 ) j = 6i + j b 3a + b = 3 ( i + 3j ) + ( 4i j ) = ( 6i + 9j ) + ( 4i j ) = ( 6 + 4 ) i + ( 9 ) j = 0i + 8j c a b = ( i + 3j ) ( 4i j ) = ( 4i + 6j ) ( 4i j ) = ( 4 4 ) i + ( 6 ( ) ) j = 7j d b + a = ( 4i j ) + ( i + 3j ) = ( 8i j ) + ( i + 3j ) = ( 8 + ) i + ( + 3 ) j = 0i + j e 3a b = 3 ( i + 3j ) ( 4i j ) = ( 6i + 9j ) ( 8i j ) = ( 6 8 ) i + ( 9 + ) j = i + j f b 3a = ( 4i j ) 3 ( i + 3j ) = ( 4i j ) ( 6i + 9j ) = ( 4 6 ) i + ( 9 ) j = i 0j g 4b a = 4 ( 4i j ) ( i + 3j ) = ( 6i 4j ) ( i + 3j ) = ( 6 ) i + ( 4 + ( 3 ) ) j = 4i 7j h a 3b = ( i + 3j ) 3 ( 4i j ) = ( 4i + 6j ) ( i 3j ) = ( 4 ) i + ( 6 ( 3 ) ) j = 8i + 9j
file://c:\users\buba\kaz\ouba\m_6_d_.html Page of Exercise D, Question Find the magnitude of each of these vectors. a 3i + 4j b 6i 8j c 5i + j d i + 4j e 3i 5j f 4i + 7j g 3i + 5j h 4i j a 3i + 4j = \ 3 + 4 = \ 9 + 6 = \ 5 = 5 b 6i 8j = \ 6 + 8 = \ 36 + 64 = \ 00 = 0 c 5i + j = \ 5 + = \ 5 + 44 = \ 69 = 3 d i + 4j = \ + 4 = \ 4 + 6 = \ 0 = 4.47 ( 3 s.f. ) e 3i 5j = \ 3 + 5 = \ 9 + 5 = \ 34 = 5.83 ( 3 s.f. ) f 4i + 7j = \ 4 + 7 = \ 6 + 49 = \ 65 = 8.06 ( 3 s.f. ) g 3i + 5j = \ 3 + 5 = \ 9 + 5 = \ 34 = 5.83 ( 3 s.f. ) h 4i j = \ 4 + = \ 6 + = \ 7 = 4. ( 3 s.f. )
file://c:\users\buba\kaz\ouba\m_6_d_3.html Page of Exercise D, Question 3 Find the angle that each of these vectors makes with the positive x-axis. a 3i + 4j b 6i 8j c 5i + j d i + 4j a 3i + 4j b 6i 8j c 5i + j d i + 4j 4 arc tan ( ) 3 8 arctan ( ) 6 arctan ( ) 5 4 arctan ( ) = 53. above ( 3 s.f. ) = 53. below ( 3 s.f. ) = 67.4 above ( 3 s.f. ) 63.4 above ( 3 s.f. )
file://c:\users\buba\kaz\ouba\m_6_d_4.html Page of Exercise D, Question 4 Find the angle that each of these vectors makes with the positive y-axis. a 3i 5j b 4i + 7j c 3i + 5j d 4i j a 3i 5j b 4i + 7j c 3i + 5j d 4i j 5 90 + arctan ( ) 3 4 arctan ( ) 7 3 arctan ( ) 5 90 + arctan ( ) 4 = 90 + 59 = 49 ( 3 s.f. ) to the right = 9.7 ( 3 s.f. ) to the right = 3.0 ( 3 s.f. ) to the left = 90 + 4 = 04 ( 3 s.f. ) to the left
file://c:\users\buba\kaz\ouba\m_6_d_5.html Page of Exercise D, Question 5 Given that a = i + 5j and b = 3i j, find a λ if a +λb is parallel to the vector i, b µ if µa + b is parallel to the vector j, a a +λb = ( i + 5j ) +λ(3i j ) = ( + 3λ ) i + ( 5 λ)j Parallel to i, so 5 λ=0,λ=5. b µa + b =µ(i + 5j ) + ( 3i j ) = ( µ + 3 ) i + ( 5µ ) j Parallel to j, so µ + 3 = 0,µ= 3
file://c:\users\buba\kaz\ouba\m_6_d_6.html Page of Exercise D, Question 6 Given that c = 3i + 4j and d = i j, find a λ if c +λd is parallel to i + j, b µ if µc + d is parallel to i + 3j, c s if c sd is parallel to i + j, d t if d tc is parallel to i + 3j. a c +λd = ( 3i + 4j ) +λ(i j ) = ( 3 +λ)i+ ( 4 λ ) j Parallel to i + j, so 3 +λ=4 λ 3λ =, λ = 3 b µc + d =µ(3i + 4j ) + ( i j ) = ( 3µ + ) i + ( 4µ ) j Parallel to i + 3j, so 4µ = 3 ( 3µ + ) 4µ = 9µ + 3, 5µ = 5,µ= c c sd = ( 3i + 4j ) s ( i j ) = ( 3 s ) i + ( 4 + s ) j Parallel to i + j, so 3 s = ( 4 + s ) 3 s = 8 + 4 s, 5 = 5 s, s = d d tc = ( i j ) t ( 3i + 4j ) = ( 3t ) i + ( 4t ) j Parallel to i + 3j, so ( 4t ) = 3 ( 3t ) 4 + 8t = 3 9t, = 7t, t = 7
file://c:\users\buba\kaz\ouba\m_6_d_7.html Page of Exercise D, Question 7 In this question, the horizontal unit vectors i and j are directed due east and due north respectively. Find the magnitude and bearing of these vectors. a i + 3j b 4i j c 3i + j d i j a i + 3j = \ + 3 = \ 4 + 9 = \ 3 = 3.6 ( 3 s.f. ) arc tan = 33.7, so bearing 034 3 b 4i j = \ 4 + = \ 6 + = \ 7 = 4. ( 3 s.f. ) arc tan = 4.0, so bearing 90 + 4 = 04 4 c 3i + j = \ 3 + = \ 9 + 4 = \ 3 = 3.6 ( 3 s.f. ) arc tan = 33.7, so bearing 70 + 34 = 304 3 d i j = \ + = \ 4 + = \ 5 =.4 ( 3 s.f. ) arc tan = 6.6, so bearing 70 7 = 43
Page of Exercise E, Question Find the speed of a particle moving with these velocities: a 3i + 4j m s b 4i 7j km h c 5i + j m s d 7i + 4j cm s a Speed = 3i + 4j = \ 3 + 4 = \ 9 + 6 = \ 5 = 5 m s b Speed = 4i 7j = \ 4 + ( 7 ) = \ 576 + 49 = \ 65 = 5 km h c Speed = 5i + j = \ 5 + = \ 5 + 4 = \ 9 = 5.39 m s ( 3 s.f. ) d Speed = 7i + 4j = \ ( 7 ) + 4 = \ 49 + 6 = \ 65 = 8.06 cm s ( 3 s.f. ) file://c:\users\buba\kaz\ouba\m_6_e_.html
file://c:\users\buba\kaz\ouba\m_6_e_.html Page of Exercise E, Question Find the distance moved by a particle which travels for: a 5 hours at velocity 8i + 6j km h b 0 seconds at velocity 5i j m s c 45 minutes at velocity 6i + j km h d minutes at velocity 4i 7j cm s. a Distance = speed time = \ 8 + 6 5 = 5 \ 64 + 36 = 5 \ 00 = 50 km b Distance = speed time = \ 5 + ( ) 0 = 0 \ 5 + = 0 \ 6 = 5.0 m ( 3 s.f. ) c Distance = speed time = \ 6 + 0.75 = 0.75 \ 36 + 4 = 0.75 \ 40 = 4.74 km ( 3 s.f. ) d Distance = speed time = \ ( 4 ) + ( 7 ) 0 = 0 \ 6 + 49 = 0 \ 65 = 967 cm ( 3 s.f. )
file://c:\users\buba\kaz\ouba\m_6_e_3.html Page of Exercise E, Question 3 Find the speed and the distance travelled by a particle moving with: a velocity 3i + 4j m s for 5 seconds b velocity i + 5j m s for 3 seconds c velocity 5i j km h for 3 hours d velocity i 5j km h for 30 minutes. a Speed = \ ( 3 ) + 4 = \ 9 + 6 = \ 5 = 5 m s, Distance = 5 5 = 75 m b Speed = \ + 5 = \ 4 + 5 = \ 9 = 5.39 m s ( 3 s.f. ) Distance = 3 5.39 = 6. m ( 3 s.f. ) c Speed = \ 5 + ( ) = \ 5 + 4 = \ 9 = 5.39 km h ( 3 s.f. ) Distance = 3 5.39 = 6. km ( 3 s.f. ) d Speed = \ + ( 5 ) = \ 44 + 5 = \ 69 = 3 km h, Distance = 0.5 3 = 6.5 km
file://c:\users\buba\kaz\ouba\m_6_f_.html Page of Exercise F, Question A particle P is moving with constant velocity v m s. Initially P is at the point with position vector r. Find the position of P t seconds later if: a r 0 = 3j, v = i and t = 4, b r 0 = i j, v = j and t = 3, c r 0 = i + 4j, v = 3i + j and t = 6, d r 0 = 3i + j, v = i 3j and t = 5. a Using r = r o + vt,r = 3j + i 4 = 8i + 3j b Using r = r o + vt,r = ( i j ) + ( j ) 3 = i j 6j = i 7j c Using r = r o + vt,r = ( i + 4j ) + ( 3i + j ) 6 = i + 4j 8i + j = 7i + 6j d Using r = r o + vt,r = ( 3i + j ) + ( i 3j ) 5 = 3i + j + 0i 5j = 7i 3j
file://c:\users\buba\kaz\ouba\m_6_f_.html Page of Exercise F, Question A particle P moves with constant velocity v. Initially P is at the point with position vector a. t seconds later P is at the point with position vector b. Find v when: a a = i + 3j, b = 6i + 3j, t =, b a = 4i + j, b = 9i + 6j, t = 5, c a = 3i 5j, b = 9i + 7j, t = 3, d a = i + 7j, b = 4i 8j, t = 3, e a = 4i + j, b = i 9j, t = 4. a Using r = r o + vt, ( 6i + 3j ) = ( i + 3j ) + v,v = ( 6i + 3j ) ( i + 3j ) = 4i + 0j v = i + 5j b Using r = r o + vt, ( 9i + 6j ) = ( 4i + j ) + 5v,5v = ( 9i + 6j ) ( 4i + j ) = 5i + 5j v = i + 3j c Using r = r o + vt, ( 9i + 7j ) = ( 3i 5j ) + 3v,3v = ( 9i + 7j ) ( 3i 5j ) = 6i + j v = i + 4j d Using r = r o + vt, ( 4i 8j ) = ( i + 7j ) + 3v,3v = ( 4i 8j ) ( i + 7j ) = 6i 5j v = i 5j e Using r = r o + vt, ( i 9j ) = ( 4i + j ) + 4v,4v = ( i 9j ) ( 4i + j ) = 8i 0j v = i 5j
file://c:\users\buba\kaz\ouba\m_6_f_3.html Page of Exercise F, Question 3 A particle moving with speed vm s in direction d has velocity vector v. Find v for these. a v = 0, d = 3i 4j c v = 7.5, d = 6i + 8j e v = \ 3, d = i + 3j g v = \ 60, d = 4i j b v = 5, d = 4i + 3j d v = 5\, d = i + j f v = \ 68, d = 3i 5j h v = 5, d = i + j a d = \ 3 + ( 4 ) = \ 5 = 5,0 5 =,v = ( 3i 4j ) = 6i 8j b d = \ ( 4 ) + 3 = \ 5 = 5,5 5 = 3,v = 3 ( 4i + 3j ) = i + 9j c d = \ ( 6 ) + 8 = \ 00 = 0,7.5 0 =,v = 6i + 8j = 4.5i + 6j d d = \ + = \,5\ \ = 5,v = 5 ( i + j ) = 5i + 5j e d = \ ( ) + 3 = \ 3,\ 3 \ 3 =,v = ( i + 3j ) = 4i + 6j f d = \ 3 + ( 5 ) = \ 34,\ 68 \ 34 = \,v = ( 3i 5j ) = 3 i 5 j g d = \ ( 4 ) + ( ) = \ 0,\ 60 \ 0 = \ 3,v = 3 ( 4i j ) = 4 3i 3j h d = \ ( ) + = \ 5,5 \ 5 = 3\ 5,v = 3 5( i + j ) = 3 5i + 6 5j 3 4 3 4
file://c:\users\buba\kaz\ouba\m_6_f_4.html Page of Exercise F, Question 4 A particle P starts at the point with position vector r 0. P moves with constant velocity v m s. After t seconds, P is at the point with position vector r. a Find r if r 0 = i, v = i + 3j, and t = 4. b Find r if r 0 = 3i j, v = i + j, and t = 5. c Find r 0 if r = 4i + 3j, v = i j, and t = 3. d Find r 0 if r = i + 5j, v = i + 3j, and t = 6. e Find v if r 0 = i + j, r = 8i 7j, and t = 3. f Find the speed of P if r 0 = 0i 5j, r = i + 9j, and t = 4. g Find t if r 0 = 4i + j, r = i j, and v = i 3j. h Find t if r 0 = i + 3j, r = 6i 3j, and the speed of P is 4 m s. a Using r = r o + vt,r = ( i ) + ( i + 3j ) 4 = i + 4i + j = 6i + j b Using r = r o + vt,r = ( 3i j ) + ( i + j ) 5 = 3i j 0i + 5j = 7i + 4j c Using r = r o + vt, ( 4i + 3j ) = r o + ( i j ) 3,r o = ( 4i + 3j ) ( 6i 3j ) = 4i + 3j 6i + 3j = i + 6j d Using r = r o + vt, ( i + 5j ) = r o + ( i + 3j ) 6,r o = ( i + 5j ) ( i + 8j ) = i + 5j + i 8j = 0i 3j e Using r = r o + vt, ( 8i 7j ) = ( i + j ) + v 3,3v = ( 8i 7j ) ( i + j ) = 6i 9j v = i 3j f Using r = r o + vt, ( i + 9j ) = ( 0i 5j ) + v 4,4v = ( i + 9j ) ( 0i 5j ) = i + 4j v = 3i + 3.5j, speed = \ ( 3 ) + 3.5 = \.5 4.6 m s g Using r = r o + vt, ( i j ) = ( 4i + j ) + ( i 3j ) t, ( i 3j ) t = ( i j ) ( 4i + j ) = 8i j,t = 4 h Using r = r o + vt, ( 6i 3j ) = ( i + 3j ) + vt,vt = ( 6i 3j ) ( i + 3j ) = 8i 6j 4t = vt 4t = 8i 6j = \ 8 + ( 6 ) = \ 00 = 0,4t = 0,4 t =.5
file://c:\users\buba\kaz\ouba\m_6_f_5.html Page of Exercise F, Question 5 The initial velocity of a particle P moving with uniform acceleration a m s is u m s. Find the velocity and the speed of P after t seconds in these cases. a u = 5i, a = 3j, and t = 4 b u = 3i j, a = i j, and t = 3 c a = i 3j, u = j + j, and t = d t = 6, u = 3i j, and a = i e a = i + j, t = 5, and u = 3i + 4j a Using v = u + at,v = ( 5i ) + ( 3j ) 4 = 5i + j speed = \ 5 + = \ 69 = 3 m s b Using v = u + at,v = ( 3i j ) + ( i j ) 3 = 3i j + 3i 3j = 6i 5j speed = \ 6 + ( 5 ) = \ 36 + 5 = \ 6 7.8 m s c Using v = u + at,v = ( i + j ) + ( i 3j ) = i + j + 4i 6j = i 5j speed = \ + ( 5 ) = \ 4 + 5 = \ 9 5.39 m s d Using v = u + at,v = ( 3i j ) + ( i ) 6 = 3i j 6i = 3i j speed = \ ( 3 ) + ( ) = \ 9 + 4 = \ 3 3.6 m s e Using v = u + at,v = ( 3i + 4j ) + ( i + j ) 5 = 3i + 4j + 0i + 5j = 7i + 9j speed = \ 7 + 9 = \ 49 + 8 = \ 30.4 m s
file://c:\users\buba\kaz\ouba\m_6_f_6.html Page of Exercise F, Question 6 A constant force F N acts on a particle of mass 4 kg for 5 seconds. The particle was initially at rest, and after 5 seconds it has velocity 6i 8j m s. Find F. Using v = u + at, 6i 8j = a 5,a = 6i 8j 5 Using F = ma,f = 4 6i 8j = 4.8i 6.4j 5
file://c:\users\buba\kaz\ouba\m_6_f_7.html Page of Exercise F, Question 7 A force i j N acts on a particle of mass kg. If the initial velocity of the particle is i + 3j m s, find how far it moves in the first 3 seconds. Using F = ma, i j = a,a = i j Using s = ut + at,s = i + 3j 3 + i j 3 = 3i + 9j + 4 i j = 7 i + 6 j distance = \ 7 + 6 = \ 56.5 + 45.565 = \ 0.85 0. m 3 4 4 3 4
file://c:\users\buba\kaz\ouba\m_6_f_8.html Page of Exercise F, Question 8 At time t = 0, the particle P is at the point with position vector 4i, and moving with constant velocity i + j m s. A second particle Q is at the point with position vector 3j and moving with velocity v m s. After 8 seconds, the paths of P and Q meet. Find the speed of Q. Using r = r o + vt for P, r = ( 4i ) + ( i + j ) 8 = 4i + 8i + 8j = i + 8j Using r = r o + vt for Q, r = ( 3j ) + v 8 Both at the same point: i + 8j = ( 3j ) + v 8,8v = i + 8j + 3j = i + j v = i + j =.5i +.375j 8 speed = v = \.5 +.375 = \.5 +.89065.03 m s
file://c:\users\buba\kaz\ouba\m_6_f_9.html Page of Exercise F, Question 9 In questions 9 and 0 the unit vectors i and j are due east and due north respectively. At pm the coastguard spots a rowing dinghy 500 m due south of his observation point. The dinghy has constant velocity ( i + 3j ) m s. a Find, in terms of t, the position vector of the dinghy t seconds after pm. b Find the distance of the dinghy from the observation point at.05 pm. a Using r = r o + vt,r = 500j + ( i + 3j ) t = 500j + ti + 3tj = ti + ( 500 + 3t ) j b 5 minutes = 5 60 seconds = 300 seconds, r = 300i + ( 500 + 3 300 ) j = 600i + 400j distance = \ 600 + 400 = \ 360 000 + 60 000 = \ 50 000 7 m
file://c:\users\buba\kaz\ouba\m_6_f_0.html Page of Exercise F, Question 0 At noon a ferry F is 400 m due north of an observation point O moving with constant velocity ( 7i + 7j ) m s, and a speedboat S is 500 m due east of O, moving with constant velocity ( 3i + 5j ) m s. a Write down the position vectors of F and S at time t seconds after noon. b Show that F and S will collide, and find the position vector of the point of collision. a Using r = r o + vt for F,r Using r = r o + vt for S,r = 400j + ( 7i + 7j ) t = 400j + 7ti + 7tj = 7ti + ( 400 + 7t ) j = 500i + ( 3i + 5j ) t = 500i 3ti + 5tj = ( 500 3t ) i + 5tj b For F and S to collide, 7ti + ( 400 + 7t ) j = ( 500 3t ) i + 5tj, i components equal: 7t = 500 3t,0t = 500,t = 50 j components equal: 400 + 7t = 5t,400 = 8t,t = 50 Both conditions give the same value of t, so the two position vectors are equal when t = 50, i.e. F and S collide at r = 7 50i + ( 400 + 7 50 ) j = 350i + 750j.
file://c:\users\buba\kaz\ouba\m_6_f_.html Page of Exercise F, Question At 8 am two ships A and B are at r A = ( i + 3j ) km and r B = ( 5i j ) km from a fixed point P. Their velocities are v A = ( i j ) km h and v B = ( i + 4j ) km h respectively. a Write down the position vectors of A and B t hours later. b Show that t hours after 8 am the position vector of B relative to A is given by ( ( 4 3t ) i + ( 5 + 5t ) j ) km. c Show that the two ships do not collide. d Find the distance between A and B at 0 am. a Using r = r o + vt for A,r = ( i + 3j ) + ( i j ) t = ( + t ) i + ( 3 t ) j Using r = r o + vt for B,r = ( 5i j ) + ( i + 4j ) t = ( 5 t ) i + ( + 4t ) j b Using AP + PB,AB = PB PA = { ( 5 t ) i + ( + 4t ) j { { ( + t ) i + ( 3 t ) j { = ( 5 t t ) i + ( + 4t 3 + t ) j = ( 4 3t ) i + ( 5 + 5t ) j c If A and B collide, the vector AB would be zero, so 4 3t = 0 and 5 + 5t = 0, but these two equations are not consistent (t = and t ), so vector AB can never be zero, and A and B will not collide. d At 0 am, t =,AB = ( 4 3 ) i + ( 5 + 5 ) j = i + 5j, Distance = \ ( ) + 5 = \ 4 + 5 = \ 9 5.39 km
file://c:\users\buba\kaz\ouba\m_6_f_.html Page of Exercise F, Question A particle A starts at the point with position vector i + j. The initial velocity of A is ( i + j ) m s, and it has constant acceleration ( i 4j ) m s. Another particle, B, has initial velocity i m s and constant acceleration j m s. After 3 seconds the two particles collide. Find a the speeds of the two particles when they collide, b the position vector of the point where the two particles collide, c the position vector of B's starting point. a Using v = u + at and t = 3, For A: v = ( i + j ) + ( i 4j ) 3 = ( + 6 ) i + ( ) j = 5i j Speed = \ 5 + = \ 5 + For B: v = i + j 3 = i + 6j = \ 46 =. ms ( 3 s.f. ) Speed = \ + 6 = \ + 36 = \ 37 = 6.08 ms ( 3 s.f. ) b Using s = ut + at for A,s = i + j 3 + i 4j 9 = 3i + 3j + 9i 8j = 6i 5j So at the instant of the collision, A is at the point with position vector r = ( i + j ) + ( 6i 5j ) = 8i 3j c Using s = ut + at for B, s = i 3 + j 9 = 3i + 9j, so B's starting point is ( 8i 3j ) ( 3i + 9j ) = 5i j
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file://c:\users\buba\kaz\ouba\m_6_g_.html Page of Exercise G, Question A particle is in equilibrium at O under the action of three forces F, F and F 3. Find F 3 in these cases. a F = ( i + 7j ) and F = ( 3i + j ) b F = ( 3i 4j ) and F = ( i + 3j ) c F = ( 4i j ) and F = ( i 3j ) d F = ( i 3j ) and F = ( 4i + j ) a F + F + F 3 = 0 ( i + 7j ) + ( 3i + j ) + F 3 = 0 F 3 = ( i + 7j ) ( 3i + j ) = i 7j + 3i j = i 8j b F + F + F 3 = 0 ( 3i 4j ) + ( i + 3j ) + F 3 = 0 F 3 = ( 3i 4j ) ( i + 3j ) = 3i + 4j i 3j = 5i + j c F + F + F 3 = 0 ( 4i j ) + ( i 3j ) + F 3 = 0 F 3 = ( 4i j ) ( i 3j ) = 4i + j i + 3j = i + 5j d F + F + F 3 = 0 ( i 3j ) + ( 4i + j ) + F 3 = 0 F 3 = ( i 3j ) ( 4i + j ) = i + 3j 4i j = 3i + j
file://c:\users\buba\kaz\ouba\m_6_g_.html Page of Exercise G, Question For each part of Question find the magnitude of F 3 and the angle it makes with the positive x-axis. a F 3 = \ ( ) + ( 8 ) = \ + 64 = \ 65 8.06 tan θ = 8 θ = 8.9 ( 3 s.f. ) angle with Ox = 8.9 below ( 3 s.f. ) b F 3 = \ ( 5 ) + ( ) = \ 5 + = \ 6 5.0 tan θ = 5 θ =.3 ( 3 s.f. ) angle with Ox = 69 above ( 3 s.f. ) c F 3 = \ ( ) + ( 5 ) = \ 4 + 5 = \ 9 5.39 tan θ = 5 θ = 68. ( 3 s.f. ) angle with Ox = 68. above ( 3 s.f. ) d F 3 = \ ( 3 ) + ( ) = \ 9 + 4 = \ 3 3.6 tan θ = 3 θ = 33.7 ( 3 s.f. ) angle with Ox = 46 above ( 3 s.f. )
file://c:\users\buba\kaz\ouba\m_6_g_3.html Page of Exercise G, Question 3 Forces P N, Q N and R N act on a particle of m kg. Find the resultant force on the particle and the acceleration produced when a P = 3i + j, Q = i 3j, r = i + j and m =, b P = 4i 3j, Q = 3i + j, r = i j and m = 3, c P = 3i + j, Q = i 5j, r = 4i + j and m = 4, d P = i + j, Q = 6i 4j, r = 5i 3j and m =. a Resultant force = P + Q + R = ( 3i + j ) + ( i 3j ) + ( i + j ) = 6i using F = ma, 6i = a, a = 3i m s b Resultant force = P + Q + R = ( 4i 3j ) + ( 3i + j ) + ( i j ) = 3i j using F = ma, 3i j = 3 a, a = i j m s c Resultant force = P + Q + R = ( 3i + j ) + ( i 5j ) + ( 4i + j ) = 3i j using F = ma, 3i j = 4 a, a = i j m s d Resultant force = P + Q + R = ( i + j ) + ( 6i 4j ) + ( 5i 3j ) = i 6j using F = ma, i 6j = a, a = i 3j m s 3 4 3
file://c:\users\buba\kaz\ouba\m_6_g_4.html Page of Exercise G, Question 4 A particle is in equilibrium at O under the action of forces X, Y and R, as shown in the diagram. Use a triangle of forces to find the magnitude of R and the value of θ when: a X = 3 N, Y = 5 N, b X = 6 N, Y = N, c X = 5 N, Y = 4 N. a Using Pythagoras' theorem, R = \ 3 + 5 = \ 9 + 5 = \ 34 5.83 N tan θ = 5 3 θ=59.0 ( 3 s.f. ) b Using Pythagoras' theorem, R = \ 6 + = \ 36 + 4 = \ 40 6.3 N tan θ = 6 θ=8.4 ( 3 s.f. )
file://c:\users\buba\kaz\ouba\m_6_g_4.html Page of c Using Pythagoras' theorem, R = \ 5 + 4 = \ 5 + 6 = \ 4 = 6.40 N 4 ( 3 s.f. ) tan θ = θ=38.7 ( 3 s.f. ) 5
file://c:\users\buba\kaz\ouba\m_6_g_5.html Page of 4 Exercise G, Question 5 The diagram shows two strings attached to a particle P, of weight W N, and to two fixed points A and B. The line AB is horizontal and P is hanging in equilibrium with BAP =α and ABP =β. The magnitude of the tension in AP is T A N, and the magnitude of the tension in BP is T B N. Use a triangle of vectors to find: a T A and T B if W = 5,α=40 and β = 50, c T A and W if T B = 5,α=40 and β = 50, e T A and α if W = 7, T B = 6 and β = 50. b T A and T B if W = 6,α=30 and β = 45, d W and T B if T A = 5,α=30 and β = 70, a The triangle of forces is a right-angled triangle T A = 5 sin 40 = 3. N ( 3 s.f. ) and T B = 5 cos 40 = 3.83 N ( 3 s.f. ) Triangle of forces
file://c:\users\buba\kaz\ouba\m_6_g_5.html Page of 4 b Triangle of forces Using the Sine rule, T B sin 60 T A = = sin 45 6 sin 75 6 sin 45 T A = = 4.39 N ( 3 s.f. ) sin 75 6 sin 60 T B = = 5.38 N ( 3 s.f. ) sin 75 c
file://c:\users\buba\kaz\ouba\m_6_g_5.html Page 3 of 4 The triangle of forces is a right-angled triangle Triangle of forces T A = 5 tan 40 = 4.0 N ( 3 s.f. ) and 5 = W cos 40, 5 W = = 6.53 N ( 3 s.f. ) cos 40 d Triangle of forces Using the Sine rule W sin 00 5 = = sin 0 T B sin 60 5 sin 00 W = = 4.4 N ( 3 s.f. ) sin 0 5 sin 60 T B = =.7 N ( 3 s.f. ) sin 0
file://c:\users\buba\kaz\ouba\m_6_g_5.html Page 4 of 4 e Triangle of forces Using the Cosine rule: T A = 6 + 7 6 7 cos 40 = 0.65 T A = 4.54 N ( 3 s.f. ) Using the Sine rule: sin 40 T A = sin ( 90 α) 6 sin ( 90 α) = 6 sin 40 4.54 = 0.848 90 α=58.,α=3.9 ( 3 s.f. )
file://c:\users\buba\kaz\ouba\m_6_h_.html Page of Exercise H, Question Three forces F, F and F 3 act on a particle. F = ( 3i + 7j ) N, F = ( i j ) N and F 3 = ( pi + qj ) N. a Given that this particle is in equilibrium, determine the value of p and the value of q. The resultant of the forces F and F is R. b Calculate, in N, the magnitude of R. c Calculate, to the nearest degree, the angle between the line of action of R and the vector j. a F + F + F 3 = 0 ( 3i + 7j ) + ( i j ) + ( pi + qj ) = 0 ( 3 + + p ) i + ( 7 + q ) j = 0 p =,q = 6 b R = F + F = ( 3i + 7j ) + ( i j ) = i + 6j c R = \ ( ) + 6 = \ 4 + 36 = \ 40 = 6.3 N tan θ = 6 θ = 8
file://c:\users\buba\kaz\ouba\m_6_h_.html Page of Exercise H, Question In this question, the horizontal unit vectors i and j are directed due east and north respectively. A coastguard station O monitors the movements of ships in a channel. At noon, the station's radar records two ships moving with constant speed. Ship A is at the point with position vector ( 3i + 0j ) km relative to O and has velocity ( i + j ) km h. Ship B is at the point with position vector ( 6i + j ) km and has velocity ( i + 5j ) km h. a Show that if the two ships maintain these velocities they will collide. The coastguard radios ship A and orders it to reduce its speed to move with velocity ( i + j ) km h. Given that A obeys this order and maintains this new constant velocity. b find an expression for the vector AB at time t hours after noon, c find, to three significant figures, the distance between A and B at 500 hours, d find the time at which B will be due north of A. a At time t r A = ( 3 + t ) i + ( 0 + t ) j r B = ( 6 t ) i + ( + 5t ) j i components equal when 3 + t = 6 t 3t = 9,t = 3 t = 3 : r A = 3i + 6j ; r B = 3i + 6j collide b New r A = ( 3 + t ) i + ( 0 + t ) j AB = r B r A = ( 6 t ) i + ( + 5t ) j { ( 3 + t ) i + ( 0 + t ) j { = ( 6 t + 3 t ) i + ( + 5t 0 t ) j = ( 9 t ) i + ( 9 + 4t ) j c t = 3 : AB = 3i + 3j, dist. = \ ( 3 + 3 ) 4.4 km d B north of A no i component 9 t = 0 t= 9 time 630 hours
file://c:\users\buba\kaz\ouba\m_6_h_3.html Page of Exercise H, Question 3 Two ships P and Q are moving along straight lines with constant velocities. Initially P is at a point O and the position vector of Q relative to O is ( i + 6j ) km, where i and j are unit vectors directed due east and due north respectively. Ship P is moving with velocity 6i km h and ship Q is moving with velocity ( 3i + 6j ) km h. At time t hours the position vectors of P and Q relative to O are p km and q km respectively. a Find p and q in terms of t. b Calculate the distance of Q from P when t = 4. c Calculate the value of t when Q is due north of P. a p = 6ti q = ( i + 6j ) + ( 3i + 6j ) t = ( 3t ) i + ( 6 + 6t ) j b t = 4 : p = 4i,q = 30j dist. apart = \ 4 + 30 = \ 576 + 900 = \ 476 38.4 km c Q north of P i components match 6t = 3t 9t = t= 3
file://c:\users\buba\kaz\ouba\m_6_h_4.html Page of Exercise H, Question 4 A particle P moves with constant acceleration ( 3i + j ) m s. At time t seconds, its velocity is v m s. When t = 0, v = 5i 3j. a Find the value of t when P is moving parallel to the vector i. b Find the speed of P when t = 5. c Find the angle between the vector i and the direction of motion of P when t = 5. a ν = u + at : v = ( 5i 3j ) + ( 3i + j ) t = ( 5 3t ) i + ( 3 + t ) j v parallel to i 3 + t = 0 t=3 s b t = 5,v = 0i + j Speed = v = 04 0. m s c 0 Angle = arctan + 90 = 68.7 s.f.
file://c:\users\buba\kaz\ouba\m_6_h_5.html Page of Exercise H, Question 5 A particle P of mass 5 kg is moving under the action of a constant force F newtons. At t = 0, P has velocity m s. At t = 4 s, the velocity of P is ( i + 5j ) m s. Find 5i 3j a the acceleration of P in terms of i and j, b the magnitude of F. At t = 6 s, P is at the point A with position vector ( 8i + 6j ) m relative to a fixed origin O. At this instant the force F newtons is removed and P then moves with constant velocity. Two seconds after the force has been removed, P is at the point B. c Calculate the distance of B from O. ν u a a = : a = i + 5j 5i 3j = 4i + j m s t 4 b F = m a = 5 ( 4i + j ) = 0i + 0j F = \ ( 0 ) + 0 = \ 500.4 N c t = 6,v = u + at v= ( 5i 3j ) + ( 4i + j ) 6 = 5i 3j 4i + j = 9i + 9j At B : r = r 0 + vt r= ( 8i + 6j ) + ( 9i + 9j ) = 0i + 4j Distance OB = \ ( 0 ) + 4 = \ 00 + 576 = \ 676 = 6 m
file://c:\users\buba\kaz\ouba\m_6_h_6.html Page of Exercise H, Question 6 In this question the vectors i and j are horizontal unit vectors in the directions due east and due north respectively. Two boats A and B are moving with constant velocities. Boat A moves with velocity 6i km h. Boat B moves with velocity ( 3i + 5j ) km h. a Find the bearing on which B is moving. At noon, A is at point O and B is 0 km due south of O. At time t hours after noon, the position vectors of A and B relative to O are a km and b km respectively. b Find expressions for a and b in terms of t, giving your answer in the form pi + qj. c Find the time when A is due east of B. At time t hours after noon, the distance between A and B is dkm. By finding an expression for AB, d show that d = 34t 00t + 00. At noon, the boats are 0 km apart. e Find the time after noon at which the boats are again 0 km apart. a 3 tan θ = θ=3, bearing is 03 5 b a b = 6ti = 3ti + ( 0 + 5t ) j c A due east of B j components match 0 + 5t = 0
file://c:\users\buba\kaz\ouba\m_6_h_6.html Page of t = 400 hours d AB = b a = { 3ti + ( 0 + 5t ) j { 6ti = 3ti + ( 0 + 5t ) j d = b a = ( 3t ) + ( 0 + 5t ) = 9t + 00 00t + 5t = 34t 00t + 00, as required. e d = 0 d = 00 34t 00t = 0 t = 0 or 00 34 time is 456 hours =.94
file://c:\users\buba\kaz\ouba\m_6_h_7.html Page of Exercise H, Question 7 A small boat S, drifting in the sea, is modelled as a particle moving in a straight line at constant speed. When first sighted at 0900, S is at a point with position vector ( i 4j ) km relative to a fixed origin O, where i and j are unit vectors due east and due north respectively. At 0940, S is at the point with position vector ( 4i 6j ) km. At time t hours after 0900, S is at the point with position vector s km. a Calculate the bearing on which S is drifting. b Find an expression for s in terms of t. At 00 a motor boat M leaves O and travels with constant velocity ( pi + qj ) km h. c Given that M intercepts S at 30, calculate the value of p and the value of q. a ( Direction of v ) = ( 4i 6j ) ( i 4j ) = 6i j tan θ = θ=8.43 3 Bearing = 90 +θ=08 b Expressing v in kmh,v = 6i j = 9i 3j r = r 0 + vt s= ( i 4j ) + t ( 9i 3j ) = ( + 9t ) i + ( 4 3t ) j 60 40 c At 30, t =.5,s = ( + 9.5 ) i + ( 4 3.5 ) j = 0.5i.5j M has been travelling for 30 minutes m=0.5 ( pi + qj ) s = m p=4,q = 3
file://c:\users\buba\kaz\ouba\m_6_h_8.html Page of Exercise H, Question 8 A particle P moves in a horizontal plane. The acceleration of P is ( i + 3j ) m s. At time t = 0, the velocity of P is ( 3i j ) m s. a Find, to the nearest degree, the angle between the vector j and the direction of motion of P when t = 0. At time t seconds, the velocity of P is v m s. Find b an expression for v in terms of t, in the form ai + bj, c the speed of P when t = 4, d the time when P is moving parallel to i + j. a tan θ = θ = 33.7 3 angle between v and j = 90 + 33.7 4 b v = u + at : v = 3i j + ( i + 3j ) t = ( 3 t ) i + ( + 3t ) j c t = 4,v = ( 3 4 ) i + ( + 3 4 ) j = 5i + 0j speed = \ ( 5 ) + 0 = \ 5 + 00 = \ 5. m s d v parallel to i + j i and j components must be equal 3 t = + 3t 5t = 5,t =
file://c:\users\buba\kaz\ouba\m_6_h_9.html Page of Exercise H, Question 9 In this question, the unit vectors i and j are horizontal vectors due east and north respectively. At time t = 0 a football player kicks a ball from the point A with position vector ( 3i + j ) m on a horizontal football field. The motion of the ball is modelled as that of a particle moving horizontally with constant velocity ( 4i + 9j ) m s. Find a the speed of the ball, b the position vector of the ball after t seconds. The point B on the field has position vector ( 9i + j ) m. c Find the time when the ball is due north of B. At time t = 0, another player starts running due north from B and moves with constant speed v m s. d Given that he intercepts the ball, find the value of v. a v = 4i + 9j speed of ball = ( 4 + 9 ) 9.85 m s b r = r 0 + vt position vector of the ball after t seconds = ( 3i + j ) + ( 4i + 9j ) t = ( 3 + 4t ) i + ( + 9t ) j c North of B when i components same, i.e. 3 + 4t = 9,4t = 6,t = 6.5 s d When t = 6.5, position vector of the ball = ( 3 + 4 6.5 ) i + ( + 9 6.5 ) j = 9i + 60.5j j component = 60.5 Distance travelled by nd player = 60.5 = 48.5 m Speed = 48.5 6.5 7.46 m s
file://c:\users\buba\kaz\ouba\m_6_h_0.html Page of Exercise H, Question 0 Two ships P and Q are travelling at night with constant velocities. At midnight, P is at the point with position vector ( 0i + 5j ) km relative to a fixed origin O. At the same time, Q is at the point with position vector ( 6i + 6j ) km. Three hours later, P is at the point with position vector ( 5i + 4j ) km. The ship Q travels with velocity i km h. At time t hours after midnight, the position vectors of P and Q are p km and q km respectively. Find a the velocity of P in terms of i and j, b expressions for p and q in terms of t, i and j. At time t hours after midnight, the distance between P and Q is d km. c By finding an expression for PQ, show that d = 58t 430t + 797 Weather conditions are such that an observer on P can only see the lights on Q when the distance between P and Q is 3 km or less. d Given that when t = the lights on Q move into sight of the observer, find the time, to the nearest minute, at which the lights on Q move out of sight of the observer. a r = r 0 + vt v P = { ( 5i + 4j ) ( 0i + 5j ) { / 3 = ( 5i + 3j ) km h b r = r 0 + vt after t hours, p = ( 0i + 5j ) + ( 5i + 3j ) t = ( 0 + 5t ) i + ( 5 + 3t ) j q = ( 6i + 6j ) + i t = ( 6 + t ) i + 6j c PQ = q p = { ( 6 + t ) i + 6j { { ( 0 + 5t ) i + ( 5 + 3t ) j { = ( 6 + 7t ) i + ( 3t ) j d = ( 6 + 7t ) + ( 3t ) = 676 364t + 49t + 66t + 9t = 58t 430t + 797, as required d d = 3 58t 430t + 797 = 69,58t 430t + 68 = 0 We are given that t = is one solution, so we know that ( t ) is a factor ( t ) ( 58t 34 ) = 0 t = 34 58 5.4 hours, so time 055 to the nearest minute