Kinematics Practice Set: NEB 11 Physics

1. Round trip

A student walks 90 m east and 90 m west in 120 s. Find distance, displacement, average speed and average velocity.

Solution: Distance = 180 m; displacement = 0. Average speed = 180/120 = 1.5 m s⁻¹. Average velocity = 0/120 = 0. Returning to the start does not erase travelled distance.

2. Position and origin

An object moves from x = −4 m to x = 7 m. Find displacement.

Solution: Δx = x_f − x_i = 7 − (−4) = +11 m. It crosses the origin, but displacement depends only on endpoints.

3. Is it speeding up?

Decide for (a) v positive, a negative; (b) v negative, a negative.

Solution: Opposite signs in (a) reduce speed. Same negative direction in (b) increases the magnitude of negative velocity, so speed grows. The sign of acceleration alone does not decide.

4. Position–time slope

Position increases linearly from 2 m at t = 0 to 22 m at t = 5 s. Find velocity.

Solution: The straight-line slope is (22−2)/(5−0) = 4 m s⁻¹. Initial position is 2 m but does not change the velocity.

5. Velocity–time trapezium

Velocity rises uniformly from 3 to 11 m s⁻¹ in 4 s. Find acceleration and displacement.

Solution: a = (11−3)/4 = 2 m s⁻². Displacement = ½(3+11)×4 = 28 m. The area unit (m s⁻¹)s = m confirms displacement.

6. Direction reversal

A v–t graph shows +6 m s⁻¹ for 5 s then −2 m s⁻¹ for 3 s. Find displacement and distance.

Solution: Signed displacement = 6×5 + (−2)×3 = 24 m. Distance = 30 + 6 = 36 m. The negative interval reverses direction but contributes positive path length.

7. Acceleration–time area

Acceleration is 3 m s⁻² for 4 s. Initial velocity is −2 m s⁻¹. Find final velocity.

Solution: Area under a–t gives Δv = 3×4 = 12 m s⁻¹. Final v = −2 + 12 = +10 m s⁻¹. The object passes through rest and reverses during the interval.

8. Starting from rest

A bus accelerates uniformly from rest at 1.8 m s⁻² for 10 s. Find final velocity and displacement.

Solution: v = u+at = 0+1.8×10 = 18 m s⁻¹. s = ut+½at² = 0+0.5×1.8×100 = 90 m.

Check: Average velocity is 9 m s⁻¹; 9×10 = 90 m.

9. Braking distance

A scooter moving at 16 m s⁻¹ stops with constant acceleration −4.0 m s⁻². Find time and displacement.

Solution: 0 = 16−4t gives t = 4.0 s. Using v² = u²+2as: 0 = 256−8s, so s = 32 m.

Check: Average velocity 8 m s⁻¹ for 4 s gives 32 m.

10. No-time equation

A train increases velocity from 8 to 20 m s⁻¹ over 168 m with constant acceleration. Find acceleration.

Solution: v² = u²+2as gives 400 = 64 + 336a, so a = 1.0 m s⁻².

Unit check: (m² s⁻²)/m = m s⁻².

11. Dropped object

A stone is released from rest. Using g = 9.8 m s⁻² and downward positive, find velocity and distance after 2.0 s.

Solution: v = 0+9.8×2.0 = 19.6 m s⁻¹ downward. s = ½×9.8×2.0² = 19.6 m downward.

Model note: This neglects air resistance. Use the value of g prescribed by the current question or course.

12. Vertical throw

A ball is thrown upward at 19.6 m s⁻¹. Take upward positive and g = 9.8 m s⁻². Find time to highest point and maximum rise.

Solution: At the top v = 0 but a = −9.8 m s⁻². From 0 = 19.6−9.8t, t = 2.0 s. From 0 = 19.6²−2×9.8s, s = 19.6 m.

Check: Acceleration remains downward at the top; only instantaneous vertical velocity is zero.

13. Launch velocity components

A ball is launched at 20 m s⁻¹, 30° above horizontal. Find initial components.

Solution: u_x = 20 cos 30° ≈ 17.3 m s⁻¹; u_y = 20 sin 30° = 10.0 m s⁻¹. In the ideal model horizontal acceleration is zero and vertical acceleration is downward.

Check: √(17.3²+10²) ≈ 20 m s⁻¹. Both components share the same flight time.