Constant acceleration 9E

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Amberly Terry
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1 Constant acceleration 9E a Take downwards as the positive direction. s = 8, = 0, a =, t =? s = t + at 8 = 0 t+ t = t 8 t = =.4 (to s.f.) The time taken for the diver to hit the water is.4 s. b v = + as v = = v = = (to s.f.) When the diver hits the water, he is travelling at m s. Take pwards as the positive direction. = 0, a =, s = 0, t =? 0 0 (0 ), 0 = t t = t t t t = = 4. (to s.f.) 0 The time of flight of the particle is 4. s. Take downwards as the positive direction. = 8, a =, t =.6, s =? s t at = + = + = (to s.f.) The height of the tower is 4 m. 4 a Take pwards as the positive direction. = 4, a=, v= 0, s =? v = + as = s s = = 9 (to s.f.) The greatest height reached by the pebble above the point of projection is 9 m. Pearson Edcation Ltd 07. Copying permitted for prchasing instittion only. This material is not copyright free.

2 4 b = 4, a =, v= 0, t =? 0 = 4 t t = =.4 (to s.f.) 4 The time taken to reach the greatest height is.4 s. 5 a Take pwards as the positive direction. = 8, a=, s = 5, v=? v = + as = 8 5 = 0 v = 0 = ± 5.5 (to s.f.) The speed of the ball when it is 5 m above its point of projection is 5.5 m s. b = 8, a =, s = 4, v=? v = + as = 8 + ( ) ( 4) = = 40.4 v = 40. = 0 (to s.f.) The speed with which the ball hits the grond is 0 m s. 6 a Take downwards as the positive direction. s = 80, = 4, a =, v=? = + v as = + = v = 584 = 40 (to s.f.) The speed with which P hits the grond is 40 m s. b = 4, a =, v= 584, t =? 584 = 4 + t t = =.7 (to s.f.) The time P takes to reach the grond is.7 s. Pearson Edcation Ltd 07. Copying permitted for prchasing instittion only. This material is not copyright free.

3 7 a Take pwards as the positive direction. v= 0, a=, t = 5, =? 0 = 5 = 5 0 = 9 The speed of projection of P is 9 m s. b = 9, v= 0, a =, s =? v = + as 0 = 9 s 5 s = = 78 (to s.f.) The greatest height above X attained by P dring its motion is 78 m. 8 Take pwards as the positive direction. =, t = 4.5, a =, s =? s t at = + = = (to s.f.) The height above the grond from which the ball was thrown is 4.7 m. 9 a Take pwards as the positive direction. s =, = 6, a=, t =? = 6t t t 6t 0 =, so sing the qadratic formla, ± t = () ( 6) ( 6) 4 () ( ) t =.4 (to s.f.) as we may discont the negative answer. The time of flight of the stone is.4 s. Pearson Edcation Ltd 07. Copying permitted for prchasing instittion only. This material is not copyright free.

4 9 b = 6, v= 0, a =, s =? v = + as 0 6 = s 6 s = = (to s.f.) The total distance travelled by the stone is ( + ) m = 9 m. 0 Take pwards as the positive direction. = 4.5, a =, s =, t =? = 4.5t t t t+ = Using the qadratic formla, t = ± -(-4.5) ( 4.5) 4 () () () =. or.9 The difference between these times is (.9.) s =.8 s The total time for which the particle is m or more above its point of projection is.8 s. a Take pwards as the positive direction. v=, a =, t =, =? = = 9.6 = 9.6 = 9.4 = 9 (to s.f.) Pearson Edcation Ltd 07. Copying permitted for prchasing instittion only. This material is not copyright free. 4

5 b = 9.4, s = 0, a=, t =? 0 = 9.4t t = t(9.4 t), t 0 t = = The time from the instant that the particle leaves O to the instant that it retrns to O is 6 s. For, take downwards as the positive direction, = 5t + t For B, take pwards as the positive direction, = 8t t B s + s = 46 B t+ t + t t = (5 ) (8 ) 46 t = 46 t = Sbstitte t = into s = 5t+ t s = + = = (to s.f.) The distance of the point where and B collide from the point where was thrown is 0 m. a Find the speed, say, immediately before the ball strikes the floor. = 0, a =, s = 0, v= v as = + = = 96 = 96 = 4 The speed of the first rebond, say, is given by = = 4 = Find the maximm height, h say, reached after the first rebond. = 0.5, v= 0, a =, s = h v = + as = h Pearson Edcation Ltd 07. Copying permitted for prchasing instittion only. This material is not copyright free. 5

6 a 0.5 h = = 5.6 (to s.f.) The greatest height above the floor reached by the ball the first time it rebonds is 5.6 m. b Immediately before the ball strikes the floor for the second time, its speed is again = 0.5 by symmetry. The speed of the second rebond, say, is given by = 4 = = Find the maximm height, h say, reached after the second rebond. = 7.875, v= 0, a =, s = h v = + as h = h = =. (to s.f.) The greatest height above the floor reached by the ball the second time it rebonds is. m. Challenge a Take pwards as the positive direction. For P, gives s = t t P For Q, Q has been moving for less second than P, so s = 0( t ) ( t ) Q t the point of collision s P = s Q t t = 0( t ) ( t ) = t t + t 0 0 = 7.8t t = =.4 (to s.f.) 7.8 The time between the instant when P is projected and the instant when P and Q collide is.4 s. Pearson Edcation Ltd 07. Copying permitted for prchasing instittion only. This material is not copyright free. 6

7 b Sbstitte t into s = t t from part a P s = t t = P (to s.f.) The distance of the point where P and Q collide from O is 7. m. Take downwards as positive. For st stone: = 0, t = t, a =, s = h h= 0 t + t = t For nd stone: = 5, t = t, a =, s = h 5( ) ( ( ) ) h= t + t = 5t 50 + ( t 4t + 4)) = 5t 50 + t 9.6t = t + 5.4t 0.4 Sbstitting for h from information for first stone: t = t + 5.4t = 5.4t 0.4 t = = Ptting this vale into eqation for first stone: h = 5.69 =.69 = 55 (to s.f.) The height of the bilding is 55 m. Pearson Edcation Ltd 07. Copying permitted for prchasing instittion only. This material is not copyright free. 7

Constant acceleration, Mixed Exercise 9

Constant acceleration, Mixed Exercise 9 a 45 000 45 km h = m s 3600 =.5 m s 3 min = 80 s b s= ( a+ bh ) = (60 + 80).5 = 5 a The distance from A to B is 5 m. b s= ( a+ bh ) 5 570 = (3 + 3 + T ) 5 ( T +