HRW 7e Chapter 4 Page of Hallida/Resnick/Walker 7e Chapter 4 3. The initial position vector r o satisfies r r = r, which results in o o r = r r = (3.j ˆ 4.k) ˆ (.i ˆ 3.j ˆ + 6. k) ˆ =.ˆi + 6.ˆj k ˆ where the understood unit is meters. 4. We choose a coordinate sstem with origin at the clock center and +x rightward (towards the 3: position) and + upward (towards : ). (a) In unit-vector notation, we have (in centimeters) r r = i and = j. Thus, Eq. 4- gives r = r r = i ˆ j ˆ. Thus, the magnitude is given b r = ( ) + ( ) = 4 cm. (b) The angle is θ = tan = 45 or 35. We choose 35 since the desired angle is in the third quadrant. In terms of the magnitudeangle notation, one ma write r = r r = i ˆ j ˆ (4 35 ). (c) In this case, r = j and r = j, and r = j cm. Thus, r = cm. (d) The angle is given b θ = tan = 9. (e) In a full-hour sweep, the hand returns to its starting position, and the displacement is zero. (f) The corresponding angle for a full-hour sweep is also zero. 5. The average velocit is given b Eq. 4-8. The total displacement r is the sum of three displacements, each result of a (constant) velocit during a given time. We use a coordinate sstem with +x East and + North. (a) In unit-vector notation, the first displacement is given b km 4. min = 6. ˆ i = (4. km)i. ˆ h 6 min/h r
HRW 7e Chapter 4 Page of The second displacement has a magnitude of north of east. Therefore, 6. =. km, and its direction is 4 km h. min 6 min/h r =. cos(4. ) ˆi +. sin(4. ) ˆj = 5.3i ˆ+.9ˆ j in kilometers. And the third displacement is km 5. min r ˆ ˆ 3 = 6. i = ( 5. km) i. h 6 min/h The total displacement is r = r + r + r = 4.i ˆ +5.3i ˆ +.9ˆj 5.ˆi = (5.3 km) ˆi +(.9 km) ˆ j. 3 The time for the trip is (4. +. + 5.) = min, which is equivalent to.83 h. Eq. 4-8 then ields v The magnitude is 5.3 km ˆ.9 km = i + ˆj = (.9 km/h) ˆi + (7. km/h) ˆ j..83 h.83 h v = (.9) + (7.) = 7.59 km/h. (b) The angle is given b or. 5 east of due north. 7. Using Eq. 4-3 and Eq. 4-8, we have 7. θ = tan = 67.5 (north of east),.9 v (.i ˆ+ 8.j ˆ.k) ˆ (5.i ˆ 6.j ˆ +.k) ˆ = = (.7i ˆ+.4j ˆ.4k) ˆ m/s. 8. Our coordinate sstem has i pointed east and j pointed north. All distances are in kilometers, times in hours, and speeds in km/h. The first displacement is rab = 483i and the second is r BC = 966 j. (a) The net displacement is r = r + r = (483 km)i ˆ (966 km)j ˆ AC AB BC
HRW 7e Chapter 4 Page 3 of which ields 3 rac = (483) +( 966) =.8 km. (b) The angle is given b 966 tan = 63.4. 483 We observe that the angle can be alternativel expressed as 63.4 south of east, or 6.6 east of south. (c) Dividing the magnitude of r AC b the total time (.5 h) gives v 483i ˆ 966j ˆ = = 5i ˆ 49j. ˆ.5 with a magnitude v = (5) + ( 49) =48 km/h. (d) The direction of v is 6.6 east of south, same as in part (b). In magnitude-angle notation, we would have v = (48 63.4 ). (e) Assuming the AB trip was a straight one, and similarl for the BC trip, then r AB is the distance traveled during the AB trip, and r BC is the distance traveled during the BC trip. Since the average speed is the total distance divided b the total time, it equals 483 + 966 = 644 km / h. 5. 9. We appl Eq. 4- and Eq. 4-6. (a) Taking the derivative of the position vector with respect to time, we have, in SI units (m/s), d ˆ v = (i + 4t ˆj + tk) ˆ = 8tˆj + k ˆ. dt (b) Taking another derivative with respect to time leads to, in SI units (m/s ), d a = (8tˆj + k) ˆ = 8ˆj. dt. We adopt a coordinate sstem with i pointed east and j pointed north; the coordinate origin is the flagpole. With SI units understood, we translate the given information into unit-vector notation as follows:
HRW 7e Chapter 4 Page 4 of ro = 4i and vo = j r = 4j and v = i. (a) Using Eq. 4-, the displacement r is r = r r = 4 ˆi+4 ˆj. o with a magnitude r = ( 4) + (4) = 56.6 m. (b) The direction of r is 4 θ = tan = tan = 45 or 35. x 4 Since the desired angle is in the second quadrant, we pick 35 ( 45 north of due west). Note that the displacement can be written as r = r r o = 56.6 35 in terms of the magnitudeangle notation. ( ) (c) The magnitude of is simpl the magnitude of the displacement divided b the time ( t = v 3 s). Thus, the average velocit has magnitude 56.6/3 =.89 m/s. (d) Eq. 4-8 shows that v points in the same direction as r, i.e, 35 ( 45 north of due west). (e) Using Eq. 4-5, we have a v vo = =. 333i +.333j t in SI units. The magnitude of the average acceleration vector is therefore. 333 =. 47 m/s. (f) The direction of a is.333 θ = tan = 45 or 35..333 Since the desired angle is now in the first quadrant, we choose 45, and a east. points north of due. In parts (b) and (c), we use Eq. 4- and Eq. 4-6. For part (d), we find the direction of the velocit computed in part (b), since that represents the asked-for tangent line. (a) Plugging into the given expression, we obtain
HRW 7e Chapter 4 Page 5 of r [.(8) 5.()]i ˆ+ [6. 7.(6)] ˆj 6.ˆ = = = i 6ˆj t. in meters. (b) Taking the derivative of the given expression produces vt t ˆ t 3 ( ) = (6. 5.) i 8. j ˆ where we have written v(t) to emphasize its dependence on time. This becomes, at t =. s, v = (9.ˆi 4ˆj) m/s. (c) Differentiating the vt () found above, with respect to t produces.tˆi 84.t ˆj, which ields a =(4.ˆi 336ˆj) m/s at t =. s. (d) The angle of v, measured from +x, is either 4 tan = 85. or 94.8 9. where we settle on the first choice ( 85., which is equivalent to 75 measured counterclockwise from the +x axis) since the signs of its components impl that it is in the fourth quadrant. 4. We make use of Eq. 4-6. (a) The acceleration as a function of time is dv d a = (( 6.t 4.t ) ) ˆ i + 8. ˆ j ( 6. 8. ) i dt = dt = t ˆ in SI units. Specificall, we find the acceleration vector at t = 3. s to be 6. 8.(3.) ˆi = ( 8 m/s )i. ˆ ( ) b (b) The equation is a = 6. 8. tgi= ; we find t =.75 s. (c) Since the component of the velocit, v = 8. m/s, is never zero, the velocit cannot vanish. (d) Since speed is the magnitude of the velocit, we have in SI units (m/s). We solve for t as follows: v = v = ( 6.t 4.t ) + ( 8.) =
HRW 7e Chapter 4 Page 6 of ( t t ) ( t t ) squaring 6. 4. + 64 = rearranging 6. 4. = 36 taking square root 6. 4. 6. rearranging 4. 6. 6. using quadratic formula t t t = ± t± = t = ( )( ) ( ) 6. ± 36 4 4. ± 6. 8. where the requirement of a real positive result leads to the unique answer: t =. s. 5. Constant acceleration in both directions (x and ) allows us to use Table - for the motion along each direction. This can be handled individuall (for x and ) or together with the unitvector notation (for r). Where units are not shown, SI units are to be understood. (a) The velocit of the particle at an time t is given b v = v + at, where v is the initial velocit and a is the (constant) acceleration. The x component is vx = v x + a x t = 3..t, and the component is v = v + a t =.5t since v =. When the particle reaches its maximum x coordinate at t = t m, we must have v x =. Therefore, 3..t m = or t m = 3. s. The component of the velocit at this time is this is the onl nonzero component of v at t m. v =.5(3.) =.5 m/s; (b) Since it started at the origin, the coordinates of the particle at an time t are given b r = v t + at. At t = t m this becomes ( )( ) ( )( ) r = 3.i ˆ 3. +.ˆi.5ˆj 3. = (4.5ˆi.5ˆ j) m. 7. (a) From Eq. 4- (with θ = ), the time of flight is h (45.) t = = = 3.3 s. g 9.8 (b) The horizontal distance traveled is given b Eq. 4-: (c) And from Eq. 4-3, we find x = vt = ( 5)( 3. 3) = 758 m. v = gt = ( 98. )( 33. ) = 97. m/s.
HRW 7e Chapter 4 Page 7 of 8. We use Eq. 4-6 R ( 9.5m/s ) v v = sin = = = 9.9 m 9.m max θ g g 9.8m/s max to compare with Powell s long jump; the difference from R max is onl R =(9. 8.95) =.59 m.. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4- are directl applicable. The initial velocit is horizontal so that v = and v = v = ms. (a) With the origin at the initial point (where the dart leaves the thrower s hand), the coordinate of the dart is given b = (b) From x = v t we obtain x = ()(.9) =.9 m. gt, so that with = PQ we have PQ = 98. 9. = 8.. (a) Using the same coordinate sstem assumed in Eq. 4-, we solve for = h: h= + vsinθt gt which ields h = 5.8 m for =, v = 4. m/s, θ = 6. and t = 5.5 s. x b gb g m. (b) The horizontal motion is stead, so v x = v x = v cos θ, but the vertical component of velocit varies according to Eq. 4-3. Thus, the speed at impact is ( θ ) ( θ ) v= v cos + v sin gt = 7.4 m/s. (c) We use Eq. 4-4 with v = and = H: b g m. vsinθ H = = 67. 5 g 5. The initial velocit has no vertical component onl an x component equal to +. m/s. Also, = +. m if the water surface is established as =. (a) x x = v x t readil ields x x =.6 m. (b) Using = v t gt, we obtain = 6.86 m when t =.8 s and v =. (c) Using the fact that = and =., the equation = v t gt (.) / 9.8.43 s t = =. During this time, the x-displacement of the diver is x x = (. m/s)(.43 s) =.86 m. leads to
HRW 7e Chapter 4 Page 8 of 3. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4- are directl applicable. The coordinate origin is at the release point (the initial position for the ball as it begins projectile motion in the sense of 4-5), and we let θ be the angle of throw (shown in the figure). Since the horizontal component of the velocit of the ball is v x = v cos 4., the time it takes for the ball to hit the wall is (a) The vertical distance is x. t = = =.5 s. v 5. cos 4. x = v θ t gt = = ( sin ) (5.sin 4. )(.5) (9.8)(.5). m. (b) The horizontal component of the velocit when it strikes the wall does not change from its initial value: v x = v cos 4. = 9. m/s. (c) The vertical component becomes (using Eq. 4-3) v = v sinθ gt= 5. sin 4. (9.8)(.5) = 4.8 m/s. (d) Since v > when the ball hits the wall, it has not reached the highest point et. 34. In this projectile motion problem, we have v = v x = constant, and what is plotted is v= v + v We infer from the plot that at t =.5 s, the ball reaches its maximum height, where x. v =. Therefore, we infer from the graph that v x = 9 m/s. (a) During t = 5 s, the horizontal motion is x x = v x t = 95 m. 9 v 3 m/s (b) Since + = (the first point on the graph), we find v = 4.5 m/s. Thus, with t =.5 s, we can use max = v t gt or v = = v g b max g, or max ( = v + v )t to solve. Here we will use the latter: max = ( v + v ) t max = ( + 45. )( 5. ) = 3 m where we have taken = as the ground level. 44. The magnitude of the acceleration is
HRW 7e Chapter 4 Page 9 of b g.. v m/s a = = = 4m/s r 5 m 45. (a) Since the wheel completes 5 turns each minute, its period is one-fifth of a minute, or s. (b) The magnitude of the centripetal acceleration is given b a = v /R, where R is the radius of the wheel, and v is the speed of the passenger. Since the passenger goes a distance πr for each revolution, his speed is b g. π 5m v = = 785m/s s and his centripetal acceleration is b g 785. m/s a = = 4. m/s. 5 m (c) When the passenger is at the highest point, his centripetal acceleration is downward, toward the center of the orbit. (d) At the lowest point, the centripetal acceleration is a = 4. m/s, same as part (b). (e) The direction is up, toward the center of the orbit. 53. To calculate the centripetal acceleration of the stone, we need to know its speed during its circular motion (this is also its initial speed when it flies off). We use the kinematic equations of projectile motion (discussed in 4-6) to find that speed. Taking the + direction to be upward and placing the origin at the point where the stone leaves its circular orbit, then the coordinates of the stone during its motion as a projectile are given b x = v t and = gt (since v = ). It hits the ground at x = m and =. m. Formall solving the second equation for the time, we obtain t = /g, which we substitute into the first equation: v b g b g g 98. m/s = x = m. m = 5. 7 m/s. Therefore, the magnitude of the centripetal acceleration is b g v 57. m/s a = = = 6 m/s. r 5. m 55. We use Eq. 4-5 first using velocities relative to the truck (subscript t) and then using velocities relative to the ground (subscript g). We work with SI units, so km / h 5.6 m / s, 3 km / h 8.3 m / s, and 45 km / h.5 m / s. We choose east as the + i direction.
HRW 7e Chapter 4 Page of (a) The velocit of the cheetah (subscript c) at the end of the. s interval is (from Eq. 4-44) v = v v =.5 ˆi ( 5.6 ˆi) = (8. m/s) ˆi c t c g t g relative to the truck. Since the velocit of the cheetah relative to the truck at the beginning of the. s interval is ( 8.3 m/s)i ˆ, the (average) acceleration vector relative to the cameraman (in the truck) is or a = 3 m/s. a 8. ˆi ( 8.3 ˆi) = = (3 m/s ) ˆi,. (b) The direction of a is +i ˆ, or eastward. (c) The velocit of the cheetah at the start of the. s interval is (from Eq. 4-44) v = v + v = ( 8.3 ˆi) + ( 5.6 ˆi) = ( 3.9 m/s) ˆ i c g c t t g relative to the ground. The (average) acceleration vector relative to the crew member (on the ground) is a.5 ˆi ( 3.9 ˆi) (3 m/s ) ˆ = = i, a = 3 m/s. identical to the result of part (a). (d) The direction of a is +i ˆ, or eastward. 56. We use Eq. 4-44, noting that the upstream corresponds to the +i ˆ direction. (a) The subscript b is for the boat, w is for the water, and g is for the ground. vb g = vb w + vw g = ( 4 km / h) i + ( 9 km / h) i = (5 km / h) i Thus, the magnitude is v = 5 km/h. bg (b) The direction of v bg is +x, or upstream. (c) We use the subscript c for the child, and obtain
HRW 7e Chapter 4 Page of vc g = vc b + vb g = ( 6 km / h) i + ( 5km / h) i = ( km / h) i. The magnitude is v = km/h. cg (d) The direction of is x, or downstream. v cg 59. Relative to the car the velocit of the snowflakes has a vertical component of 8. m/s and a horizontal component of 5 km/h = 3.9 m/s. The angle θ from the vertical is found from which ields θ = 6. vh 3.9 m/s tanθ = = =.74 v 8. m/s v 6. The destination is D = 8 km j^ where we orient axes so that + points north and +x points east. This takes two hours, so the (constant) velocit of the plane (relative to the ground) is v pg = 4 km/h j^. This must be the vector sum of the plane s velocit with respect to the air which has (x,) components (5cos7º, 5sin7º) and the velocit of the air (wind) relative to the ground v ag. Thus, 4 j^ = 5cos7º i^ + 5sin7º j^ + v ag v ag = 7i^ 7.j^. (a) The magnitude of v is v = ( 7) + ( 7.) = 85 km/h. ag ag (b) The direction of v ag is 7. θ = tan =.3 (south of west). 7 69. Since v = v g, and v= at the target, we obtain ( )( ) v = 9.8 5. = 9.9 m/s (a) Since v sin θ = v, with v =. m/s, we find θ = 55.6. (b) Now, v = v gt gives t = 9.9/9.8 =. s. Thus, x = (v cos θ )t = 6.85 m. (c) The velocit at the target has onl the v x component, which is equal to v x = v cos θ = 6.78 m/s. 7. The (x,) coordinates (in meters) of the points are A = (5, 5), B = (3, 45), C = (, 5), and D = (45, 45). The respective times are t A =, t B = 3 s, t C = 6 s, and t D = 9 s.
HRW 7e Chapter 4 Page of Average velocit is defined b Eq. 4-8. Each displacement r is understood to originate at point A. (a) The average velocit having the least magnitude (5./6) is for the displacement ending at point C: v =.83 m/s. (b) The direction of v is (measured counterclockwise from the +x axis). (c) The average velocit having the greatest magnitude ( 5 + 3 3 ) is for the displacement ending at point B: v =. m/s. (d) The direction of v is 97 (counterclockwise from +x) or 63 (which is equivalent to measuring 63 clockwise from the +x axis). 88. Eq. 4-34 describes an inverse proportionalit between r and a, so that a large acceleration results from a small radius. Thus, an upper limit for a corresponds to a lower limit for r. (a) The minimum turning radius of the train is given b r min b g b gc h v 6 km / h = = = 73. amax. 5 9. 8 m/s (b) The speed of the train must be reduced to no more than which is roughl 8 km/h. 3 v = amax r =. 5 9. 8. = 3 m. b gc h m/s 3. The initial velocit has magnitude v and because it is horizontal, it is equal to v x the horizontal component of velocit at impact. Thus, the speed at impact is v + v = 3v where v = gh and we have used Eq. -6 with x replaced with h = m. Squaring both sides of the first equalit and substituting from the second, we find v gh 3v + =b g which leads to gh = 4v and therefore to v = (.)( 98 )/ = 7. m/s.