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AQA AS Physics C7 On The Move Kerboodle Answers

This page contains the AQA AS Physics P7 On The Move Kerboodle Answers for revision and understanding of On The Move . This page also contains the link to the notes and video for the revision of this topic. 
 
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7.1 Speed and velocity AQA AS Physics C7 On The Move Kerboodle Answers: Page No. 119

1.Ans- a km/h 80kmh-1 b ms-1. 22ms-1 2.Ans-a 2.5 m s-1 b 3.0 ms-1 3.Ans-a 2.5 X 104 kmh-1 b 7.0 x 103 ms-1 4.Ans- a 45000m b 5.Ans- a b i 4.0 km ii 30ms-1 then 25ms-1 in the opposite direction

7.2 Acceleration AQA AS Physics C7 On The Move Kerboodle Answers: Page No. 121

1. a1.5ms-2 b A-2.5 ms-2 2.a0.45 ms-2 b . 7.9 ms-1 3 a Sketch a velocity—time graph to represent its journey. b 0.60ms-2, 0, -0.40ms-2 4 Describe how a the velocity of the object changed with time, b the acceleration of the object changed with time.

7.3 Motion along a straight line at constant acceleration AQA AS Physics C7 On The Move Kerboodle Answers : Page No. 124

1.a 2.0 ms-2 b 221m 2.a 43s b -0.93ms-2 3.a i the acceleration, 0.2 m s-2 ii the displacement. 90 m b i -0.75ms-2 ii 8.0 s d 3.0ms-1 4. a5.0ms-2 b the displacement in the first s 7.5m c the displacement in the next 2 S 18m d the average speed over the first 4 s. 6.4ms-1

7.4 Free fall AQA AS Physics C7 On The Move Kerboodle Answers : Page No. 127 1.4.0m b its speed just before hitting the water. 8.8ms-1 2. a the time taken for the spanner to hit the ground 3.2s b the speed of impact of the spanner on hitting the ground. 31 ms-1 3.3.9s ii the speed of impact of the object on hitting the water, assuming air resistance is negligible. 38ms-1 4.a the acceleration due to gravity On the Moon 1.6ms-2 b the speed of projection of the object 3.6ms-1 c 0.64m

7.5 Motion graphs AQA AS Physics C7 On The Move Kerboodle Answers : Page No. 129

1.a Calculate how long the swimmer takes to swim from i the starting end to the other end, 83 s ii back to the start from the other end. 127 s b For the swim from starltc finish, sketch i a displacement—time graph, ii a distance—time graph, iii a velocity—time graph. 2.a Calculate how long the motorcyclist has to wait before the cyclist catches up. 600 s b Cn the same axes, sketch velocity—time graphs for i the motorcyclist, ii the cyclist. 3. a Sketch a graph to show how the velocity of the object changed with time during its descent. Velocity/ m s-1 b Show that the balloon base wag 33.6 above the ground when the object fell off the base. 4.a Given 9 ms-a, calculate i the duration Of its descent, 0.61 s ii its velocity just before impact, 5.9 ms-1 iii the duration of its ascent, 0.43 s iv its velocity just after impact. 4.2 ms-1.

7.6 More calculations on motion along a straight line AQA AS Physics C7 On The Move Kerboodle Answers : Page No. 131

1.a Calculate i the time taken to reach 29 ms-1 from 4 m s- l, 52s ii its acceleration during this time. 0.49ms-2 b Calculate i the distance it travelled during deceleration, 406m ii its deceleration for the last 28s. -1.04ms-2 2.a the distance it moved up the incline 15 m b its acceleration on the incline -0.13ms-2 c its velocity and position on the incline after 0.67ms-1 downwards, 13.4m from the start 3.a Calculate i the distance travel led by the cyclist in the first 20 s, 80m ii the speed of the cyclist at the end Of this time. 8.0ms-1 b Calculate i the time taken to cover the distance of 260m after she stopped pedalling, 65s ii her deceleration during this time. -0.12ms-2 4.a Calculate i its velocity, 180ms-1 ii its height above the launch pad when its motors switched off. 2.7 km b Calculate the maximum height reached after its motors switched off, 4.4km c Calculate the velocity with which it would hit the ground if it fell from maximum height without the support of a parachute. 290ms-1 Banner 7

7.7 Projectile motion 1 AQA AS Physics C7 On The Move Kerboodle Answers : Page No. 133

1 a the velocity of the object at the ground 32ms-1 b the duration of descent 2.8s c the height of the balloon above the ground when the object hits the ground. 39m 2.a how long it takes to reach the sea 3.0s b how far it travels horizontally 49m c its impact velocity. 34ms-1 (at 62° to the horizontal) 3.a the time of flight of the dart 0.20s b its horizontal speed Of projection. 11.7ms-1 4. a the height Of the aircraft above the ground 354m b the horizontal distance travelled in this time bg i the parcel, 1020m ii the aircraft 1020m c the speed of impact of the parcel at the ground. 146ms-1

7.8 Projectile motion 2 AQA AS Physics C7 On The Move Kerboodle Answers : Page No. 135

1. a 470mm b 3.0ms-2 c 2.7ms-1 2 a 2.02s b50.5m 3 a3.5ms-1, 3.0ms-1 b150m ii 20m Banner 8

Practice questions: Page No. 136-137

1.(a) ) (b) 2 . (a) AB: uniform acceleration BC: constant velocity CD: uniform deceleration DE: stationary EF: uniform acceleration in the opposite direction (b) Displacement is equal to the area enclosed by the graphs and the time axis (c) Distance is a scalar and is represented by the total area under both the positive and negative portions of the graph whereas displacement is a vector and the areas abo’ve and below the v = 0 line are equal and therefore cancel 3.(a) Using v = u + a t gives 12 = 4 + 6.0 a ∴acceleration a = 1.3 m s-2 (b) (c) Distance travelled = area under graph = area of trapezium with vertical sides of 4 m s-1 and 12 m s-1 = 1/2 × (4 + 12) × 6.0 = 48 m 4 .(a) (i) Using v = u + a t gives 0 = 4.5 + 3600 a ∴ Acceleration a = −1.3 × 10-3 m s-2 and deceleration = 1.3 × 10-3 m s-2 (ii) (b) (c) • Gradient of graph = speed • Speed decreases giving a decreasing gradient • Gradient is zero when stationary 5.(a) i) Its acceleration Using v = u + a t gives 29 = 0 + 2.0 a ∴ Acceleration a = 14.5 m s-2 (ii) (iii) Using distance = (speed) × (time) gives s = 29 × 15 = 435 m (b) (i) Using the same axes plot the speed—time graph for the antelope during the chase. (ii) (iii) 6 (a) (b) A lower value would be obtained for g Because air resistance has a greater effect on the tennis ball Resulting in a smaller resultant downwards force on the tennis ball (c) 7 (a) Horizontal velocity remains 70 m s-1 (b) Using v = u + a t gives vertical velocity vV = 0 + (9.81 × 2.0) = 19.6 m s-1 (c) Banner 9

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