Equations of Motion – Long Answer Questions

Medium Level (Application & Explanation)

Q1. Derive the equation v = u + at for motion with uniform acceleration and explain the meaning of each term.

Answer:

Start with the definition of acceleration: acceleration a is the rate of change of velocity with time, so a = (v − u)/t.
Rearranging, we get v − u = at.
Therefore, v = u + at.
Here u is the initial velocity, the velocity at time t = 0. v is the final velocity after time t. a is the uniform acceleration, which remains constant during the motion. t is the time interval for which acceleration acts.
This equation tells us how the velocity changes linearly with time when acceleration is constant. It is used to find final speed when initial speed, acceleration, and time are known.

Q2. Explain how s = ut + (1/2)at^2 is obtained and describe a practical situation where this equation is useful.

Answer:

To derive s = ut + 1/2 at^2, start with average velocity when acceleration is uniform: average velocity = (u + v)/2.
Using v = u + at, average velocity = (u + (u + at))/2 = u + (1/2)at.
Distance s = (average velocity) × t = [u + (1/2)at]t = ut + (1/2)at^2.
Here s is displacement in time t, u initial velocity, a constant acceleration.
Practical use: When a car starts from rest and accelerates uniformly, this equation calculates the distance traveled in a given time. For example, calculating how far a sprinter runs while accelerating in the first few seconds.

Q3. Explain the physical meaning and use of v^2 − u^2 = 2as when time is not known.

Answer:

The equation v^2 − u^2 = 2as links velocity and displacement without time. It is derived by eliminating t between v = u + at and s = ut + 1/2 at^2.
Physically, it tells how the change of kinetic energy-like quantity (square of speed) depends on acceleration and distance traveled. If acceleration is positive, v > u; if decelerating (a negative), v < u.
Use case: Calculating stopping distance of a car under braking when the braking acceleration and initial speed are known but time to stop is not given. It is widely used in safety and engineering problems where time is difficult to measure.

Q4. A ball is dropped from rest from a height. Using equations of motion, explain step-by-step how to find the time taken and final speed just before hitting the ground. Assume g as acceleration.

Answer:

For a ball dropped from rest, initial velocity u = 0, acceleration a = g (downward), displacement s is the height.
To find time t: use s = ut + (1/2)at^2 → s = 0 + (1/2)gt^2. Solve t = sqrt(2s/g). This gives time to fall.
To find final speed v just before impact: use v = u + at → v = 0 + gt = g × sqrt(2s/g) = sqrt(2gs).
Thus, time depends on height and g via t = sqrt(2s/g), and final speed is v = sqrt(2gs). These formulas are direct results of uniform acceleration due to gravity.

Q5. How would you use equations of motion to determine how far a car travels while accelerating from 10 m/s to 30 m/s with constant acceleration of 2 m/s^2?

Answer:

Known: u = 10 m/s, v = 30 m/s, a = 2 m/s^2, and we need s. Time t is not required.
Use the equation v^2 − u^2 = 2as. Substitute values: (30)^2 − (10)^2 = 2 × 2 × s. That is 900 − 100 = 4s, so 800 = 4s.
Solve for s: s = 800 / 4 = 200 m.
Thus the car travels 200 m during the speed change. This method avoids calculating time and directly relates speeds to distance under uniform acceleration.

High Complexity (Analytical & Scenario-Based)

Q6. Two cars A and B move along a straight road. Car A starts from rest and accelerates uniformly at 1.5 m/s^2. Car B passes the same point 4 s later moving at 20 m/s in the same direction and continues with constant speed. Determine whether car A will catch car B. If yes, find the time and distance from the starting point where they meet.

Answer:

For Car A: u_A = 0, a_A = 1.5 m/s^2. Position after time t: s_A = 0·t + (1/2)·1.5·t^2 = 0.75 t^2. Velocity v_A = 1.5 t.
Car B passes the point at t = 4 s with speed 20 m/s and constant speed. For times t ≥ 4, position of B from A’s start: s_B = 20·(t − 4) (since it started measured from the same point at t = 4). Add zero offset because we measure from same point.
Set s_A = s_B: 0.75 t^2 = 20(t − 4). Rearranged: 0.75 t^2 − 20t + 80 = 0. Multiply by 4: 3 t^2 − 80 t + 320 = 0.
Solve quadratic: Discriminant D = 80^2 − 4·3·320 = 6400 − 3840 = 2560. sqrt(D) ≈ 50.596. t = [80 ± 50.596]/(6). Positive roots: t1 ≈ (80 − 50.596)/6 ≈ 29.404/6 ≈ 4.9007 s; t2 ≈ (80 + 50.596)/6 ≈ 130.596/6 ≈ 21.766 s.
The first time t ≈ 4.90 s is just after B passed (t>4), so car A catches B at t ≈ 4.90 s. Distance: s_A = 0.75·(4.9007)^2 ≈ 0.75·24.01 ≈ 18.01 m.
Conclusion: Yes, A catches B about 4.90 s after A starts, at approximately 18.0 m from the start point.

Q7. A cyclist moving at 8 m/s applies brakes and uniformly decelerates to rest in 5 s. Calculate the stopping distance and sketch (describe) what the velocity-time and distance-time graphs look like.

Answer:

Known: u = 8 m/s, v = 0, t = 5 s. Acceleration a = (v − u)/t = (0 − 8)/5 = −1.6 m/s^2 (deceleration).
Stopping distance s using s = ut + (1/2)at^2: s = 8·5 + 0.5·(−1.6)·25 = 40 − 20 = 20 m. So stop in 20 m.
Velocity-Time graph: a straight line sloping downward from v = 8 m/s at t = 0 to v = 0 at t = 5 s. This line is linear because acceleration is constant. The area under this line equals displacement.
Distance-Time graph: a curving upward graph starting at s = 0 with a slope equal to instantaneous velocity; slope decreases linearly and reaches zero at t = 5 s. The curve is concave (since acceleration is negative) and levels off at s = 20 m.

Q8. A rocket accelerates uniformly from 50 m to 210 m in 8 s, starting from rest. Calculate its acceleration and final velocity, and explain each step using equations of motion.

Answer:

Clarify: The rocket travels s = 210 − 50 = 160 m in t = 8 s starting from rest, so u = 0.
Use s = ut + 1/2 at^2 → 160 = 0 + 1/2 · a · 64 → 160 = 32 a → a = 160/32 = 5 m/s^2. So acceleration is 5 m/s^2.
Final velocity v = u + at = 0 + 5 × 8 = 40 m/s. So final speed is 40 m/s.
Explanation: We first used the position-time equation because it directly relates s, u, a, and t. After finding a, we applied v = u + at to find the final velocity. Both steps assume uniform acceleration and straight-line motion.

Q9. A runner accelerates uniformly from 3 m/s to 9 m/s over a distance of 84 m. Determine the acceleration and time taken. Show your reasoning using equations of motion.

Answer:

Known: u = 3 m/s, v = 9 m/s, s = 84 m. Time t and acceleration a are unknown. Use v^2 − u^2 = 2as to find a.
Compute: (9)^2 − (3)^2 = 2a·84 → 81 − 9 = 168 a → 72 = 168 a → a = 72/168 = 0.428571... ≈ 0.4286 m/s^2.
Now find time using v = u + at: 9 = 3 + (0.4286) t → t = (9 − 3)/0.4286 ≈ 6/0.4286 ≈ 14.0 s.
So acceleration ≈ 0.429 m/s^2 and time ≈ 14 s. We used the third equation to avoid time first, then the first to get t.

Q10. A skateboarder starts from rest, accelerates at 2 m/s^2 for 4 s, then immediately decelerates uniformly to stop in 3 s. Calculate the maximum speed, distance covered during acceleration, distance during deceleration, and total distance. Explain how equations of motion are combined in this two-stage motion.

Answer:

Stage 1 (acceleration): u1 = 0, a1 = 2 m/s^2, t1 = 4 s. Max speed v_max = u1 + a1 t1 = 0 + 2×4 = 8 m/s. Distance s1 = u1 t1 + 1/2 a1 t1^2 = 0 + 0.5×2×16 = 16 m.
Stage 2 (deceleration): starts with u2 = v_max = 8 m/s, final v2 = 0, t2 = 3 s. Deceleration a2 = (v2 − u2)/t2 = (0 − 8)/3 ≈ −2.6667 m/s^2. Distance s2 = u2 t2 + 1/2 a2 t2^2 = 8×3 + 0.5×(−2.6667)×9 = 24 − 12 = 12 m.
Total distance = s1 + s2 = 16 + 12 = 28 m.
Explanation: Treat each stage separately with appropriate initial speeds and accelerations. Use s = ut + 1/2 at^2 for distances and v = u + at for speeds. Combining stages yields the total motion because motion is along the same straight line.

Study tips: For all problems, clearly identify u, v, a, t, s, choose the appropriate equation, and check units. Use v^2 − u^2 = 2as when time is unknown, and use v = u + at or s = ut + 1/2 at^2 when time is given or needed.