Constant acceleration, Mixed Exercise 9
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1 Constant acceleration, Mixed Exercise 9 a km h = m s 3600 =.5 m s 3 min = 80 s b s= ( a+ bh ) = ( ).5 = 5 a The distance from A to B is 5 m. b s= ( a+ bh ) = ( T ) 5 ( T + 64) = T + 64 = = 76 5 T = = c At t = 3, s = 3 5 = 480 At t = 44, s = area of the triangle = = 570 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
2 c 3 a i Gradient of line = v u t v u a = t Rearranging: ii Shaded area is a trapezium area = t s = t b i Rearrange v u t = a Substitute into s = t v u s = a as = v u v = u + as ii Substitute into u + u + at s = t u at s = + t s = t Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
3 3 b iii Substitute u = v at into s = ( v at) t + at s = vt at 4 s= ( a+ bh ) 5 = (5 + 3) u = 9 u u = 5 = km h = m s = m s km h = m s = m s u =, v=, s = 40, a = v = u + as 0 00 ( 3 ) ( 9 ) 0 ( ) 00 ( ) = + a 40 a = = 0.65 (to s.f.) The deceleration of the car is 0.65 m s. 6 a a =.5, u = 0, t =, s = = 0.5 = = 60 OA = 60 m b The particle will turn round when v = 0 a=.5, u = 0, v= 0, s = = + v u as = s s = The total distance P travels is (80 + 0) m = 00m Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 3
4 7 u = 6, v= 5, a = 9.8, t = 5 = t 5 6 t = =.9 (to s.f.) 9.8 The ball takes.9 s to move from the top of the tower to the ground. 8 Take downwards as the positive direction. a u = 0, s = 8, a = 9.8, t = 8 = t 8 t = = 4. (to s.f.) 4.9 The time taken for the ball to reach the sea is 4.s. b u = 0, s = 8, a = 9.8, v = = + v u as = = 607. v = 607. = 40 (to s.f.) The speed at which the ball hits the sea is 40 m s. c Air resistance/wind/turbulence 9 a distance = area of triangle + area of rectangle + area of trapezium 45 = 8 u+ u+ ( u+ u) 6 = 8u+ 4u+ 9u = 4 u u = = 45 4 b The particle is moving with speed less than u m s for the first 4 s s = 4 = The distance moved with speed less than u m s is m. 0 a From O to P, u = 8, t =, v= 4, a= u = 8, t =, v= 4, a = 4 = 8 + a Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 4
5 0 a 4 8 a = = From O to Q, u = 8, t = 0, a=, v= = = 8 The speed of the train at Q is 8 m s. b From P to Q u = 4, v= 8, t = 8, s = s = t = 8 = 08 The distance from P to Q is 08 m. a s = 04, t = 8, v= 8, u = s = t u = 8 = ( u+ 8) 4 = 4u u = = 8 4 The speed of the particle at X is 8 m s b s = 04, t = 8, v= 8, a= s = vt at = a = a = =.5 3 a The acceleration of the particle is.5 ms. c From X to Z, u = 8, v= 4, a =.5, s = v = u + as = + s s = = XZ = 04.8 m Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 5
6 a Take upwards as the positive direction. u =, s = 3, a= 9.8, v= = + v u as ( 9.8) ( 3) = + = + = v = 068. = ± 33 (to s.f.) The velocity with which the pebble strikes the ground is 33 m s. The speed is 33 m s. b 40 m above the ground is 8 m above the point of projection. u =, s = 8, a= 9.8, t = 8 = t 4.9 t = t t , so using the quadratic formula, ± ± 84. t = = = 3.86, 0.43 (to 3 s.f.) The pebble is above 40 m between these times: = 3.44 (to 3 s.f.) The pebble is more than 40 m above the ground for 3.44 s. c Take upwards as the positive direction. u =, a = 9.8 v = 9.8t t = 9.8 From part a, the pebble hits the ground when v = 33. v ( 33) 54 t = = = = 5.5 (to s.f.) This is shown on the graph at point (5.5, 33) The graph crosses the t-axis when v = 0. v 0 t = = = =. (to s.f.) So t he graph passes through point (., 0) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
7 c 3 a u =, v= 3, s = 00, t = s = t = t t t 00 = = = 50 The time taken by the car to move from A to C is 50 s. b Find a first. From A to C, u =, v= 3, t = 50, a = 3 = + a 50 3 a = = From A to B, u =, s = 550, a = 0.4, v= v = u + as = + = = v 4. (to 3 s.f.) The car passes B with speed 4. m s. 4 Take upwards as the positive direction. At the top: u = 30, v= 0, a = 9.8, t = 30 0 = t t = 9.8 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7
8 4 The ball spends.4 seconds above h, thus (by symmetry). seconds rising between h and the top. 30 So it passes h. seconds earlier, at t =. =.86 (to 3 s.f.) 9.8 At h, u = 30, a= 9.8, t.86, s = ( 9.8) (to s.f.) = + = 5 a u = 0, a = 4, s = 78, v= v = u + as = + = v = 04 = 3 The speed of B when it has travelled 78 m is 3 m s. b Find time for B to reach the point 78 m from O. v= 3, u = 0, a= 4, t = = 0 + 4t t = = 3 4 For A, distance = speed time s = 30 3 = 90 The distance from O of A when B is 78 m from O is 90 m. c At time tseconds, for As, = 30t for B, = 0t + t On overtaking the distances are the same. 0 + = 30 t t t t t tt 5 = ( 5) = 0 t = 5 (at t = 0, A overtakes B) B overtakes A 5 s after passing O. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 8
9 6 a To find time decelerating: u = 34, v=, a = 3, t = 34 = 34 3t t = = 4 3 b distance = rectangle + trapezium s = 34 + ( + 34) 4 = 68 + = 80 Distance required is 80 m. 7 a b Acceleration is the gradient of a line For the first part of the journey, 3x= t = = t 3x x For the last part of the journey, x= t = t x t + T + t = T + = T = 300, as required x x x c s= ( a+ bh ) 6000 = ( T + 300) 30 = 5T Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 9
10 7 c T = = 00 5 Substitute into the result in part b: + 00 = 300 = x x 40 x = 00 = 0. d From part c, T = 00 At constant velocity, distance = velocity time = = 3000 (m) The distance travelled at a constant speed is 3 km. 0 0 e b x 0. From part, t = = = 50 Total distance travelled = 6 km (given) so halfway = 3 km = 3000 m ( ) While accelerating, distance travelled is m = 750 m. At constant velocity, the train must travel a further 50 m. distance 50 At constant velocity, time = = s = 75 s velocity 30 Time for train to reach halfway is ( ) s = 5 s Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 0
11 Challenge Find the time taken by the first ball to reach 5 m below its point of projection (5 m above the ground). Take upwards as the positive direction. u = 0, s = 5, a = 9.8, t = 5 = 0t 4.9 t = t t t = 0 ± 9.8 = 3.5 (to s.f.) As we discard the negative solution. Find the time taken by the second ball to reach 5 m below its point of projection. Take downwards as the positive direction. u = 0, s = 5, a = 9.8, t = 5 = 4.9t t =.3 (to s.f.) Combining the two results: T = =. (to s.f. using exact figures) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
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