The velocity of a car changes from 25 m s−1 north to 25 m s−1 east in 10 s. What is the magnitude of the average acceleration?
- A 3.5 m s−2
- B 0
- C 5.0 m s−2
- D 2.5 m s−2
Show answer
Answer: A
Free Kinematics questions for AS Level Physics (9702), part of Mechanics & materials. Each question has the answer and the marking points an examiner would look for.
Generate a full Kinematics worksheet →All Mechanics & materials questionsThe velocity of a car changes from 25 m s−1 north to 25 m s−1 east in 10 s. What is the magnitude of the average acceleration?
Answer: A
A body falls through air and reaches terminal velocity. Which statement is correct at terminal velocity?
Answer: B
Which term is described by this statement?
“The distance moved in a particular direction.”
Answer: C
A velocity–time graph for a body is a straight line with negative gradient crossing the time axis. What happens at the crossing?
Answer: D
A ball is thrown horizontally at 18 m s−1 from the top of a vertical cliff of height 45 m. Air resistance is negligible.
Explain why the horizontal component of the velocity of the ball remains constant.
Calculate the time taken for the ball to reach the ground.
Use your answer to (b) to calculate the horizontal distance travelled.
Calculate the speed of the ball just before it hits the ground, and the angle its velocity makes with the horizontal.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | no horizontal force acts no (resultant) horizontal force, so no horizontal acceleration B1 | 1 |
| (b) | 3.03 s s = ½gt2 with s = 45 C1 3.03 s A1 | 2 |
| (c) | 54.5 m 18 × 3.03 = 54.5 m B1 | 1 |
| (d) | 34.7 m s−1 at 58.8° vy = g t = 29.7 m s−1 C1 speed = √(182 + 29.72) = 34.7 m s−1 A1 angle = tan−1(29.7/18) = 58.8° A1 | 3 |
A ball is thrown with velocity 12 m s−1 at 50° above the horizontal from a point 10 m above flat ground. Air resistance is negligible.
Calculate the maximum height above the ground.
Calculate the time taken to reach the ground.
Calculate the horizontal distance travelled.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 14.3 m v2 = u2 + 2as with vertical component C1 14.3 m A1 | 2 |
| (b) | 2.64 s s = ut + ½at2 C1 2.64 s A1 | 2 |
| (c) | 20.4 m 20.4 m B1 | 1 |
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