Q1. Explain the meaning and use of Equation (7.5): v = u + at. How does it help in solving motion problems?
Answer:
Equation (7.5) shows how final velocity (v) changes with initial velocity (u) under uniform acceleration (a) over time (t).
It tells us that velocity changes by a fixed amount every second when acceleration is constant.
If a is positive, the object speeds up. If a is negative (retardation), the object slows down.
This equation is useful when we know u, a, t and we want v.
It is often the first step in problems where time is given or can be found.
The units must be consistent, usually in m/s for velocity and m/s² for acceleration.
Always keep the sign of acceleration correct based on direction.
Q2. A car starts with u = 12 m/s and accelerates at a = 2 m/s² for t = 8 s. Use Equation (7.6) to find the distance. Explain each step.
Answer:
We use Equation (7.6): s = ut + (1/2) a t².
Here, u = 12 m/s, a = 2 m/s², and t = 8 s.
Substitute the values: s = (12 × 8) + (1/2 × 2 × 8²).
So, s = 96 + (1 × 64) = 160 m.
This means the car covers 160 meters in 8 seconds.
The first part ut is distance due to initial speed.
The second part (1/2)at² is distance added due to acceleration.
Q3. How do you choose the correct equation of motion for a given problem? Explain with a short example.
Answer:
First, list knowns: u, v, a, t, s.
If time (t) is present, use v = u + at or s = ut + (1/2)at².
If time is missing, use 2as = v² − u².
Pick the equation that contains the unknown and the known values only.
Example: If u = 5 m/s, a = 3 m/s², t = 4 s, and s is needed, use s = ut + (1/2)at².
If u = 10 m/s, v = 30 m/s, a = 5 m/s², and s is needed, use 2as = v² − u².
This approach saves time and avoids mistakes.
Q4. A scooter moving at 25 m/s stops under a uniform deceleration of 5 m/s². Find the stopping distance using Equation (7.7). Explain the sign carefully.
Answer:
Use Equation (7.7): 2as = v² − u².
Here, u = 25 m/s, v = 0 m/s, a = −5 m/s² (negative due to braking).
Substitute: 2 × (−5) × s = 0² − 25² = −625.
So, −10s = −625, hence s = 62.5 m.
The distance is positive because distance is a scalar.
The negative acceleration shows slowing down.
This equation is best when time is not given.
Q5. What is uniform acceleration? Explain its units, sign convention, and give two simple examples from daily life.
Answer:
Uniform acceleration means the rate of change of velocity is constant.
Its unit is m/s² (meters per second squared).
If velocity increases in the chosen direction, a is positive. If it decreases, a is negative.
Example 1: A car speeding up from a traffic light has positive acceleration.
Example 2: A bike braking to stop has negative acceleration (retardation).
In vertical motion, take upward as positive, then gravity is a = −9.8 m/s² (≈ −10 m/s²).
Always use a consistent sign convention for correct answers.
High Complexity (Analysis & Scenario-Based)
Q6. A car at 20 m/s brakes uniformly to rest in 5 s. Find the acceleration and the stopping distance. Solve it in two ways and discuss signs.
Answer:
Given: u = 20 m/s, v = 0, t = 5 s.
Acceleration from (7.5): a = (v − u)/t = (0 − 20)/5 = −4 m/s².
Method 1 (using time): s = ut + (1/2)at² = (20 × 5) + 0.5 × (−4) × 25 = 100 − 50 = 50 m.
Method 2 (no time): 2as = v² − u² ⇒ 2 × (−4) × s = 0 − 400 ⇒ s = 50 m.
Both methods agree, so the stopping distance is 50 m.
The negative sign in acceleration shows slowing down.
Keep directions and signs consistent to avoid errors.
Q7. A ball is thrown upward with u = 20 m/s. Take a = −10 m/s². Find time to reach the highest point and the maximum height. Explain why acceleration is negative.
Answer:
At the highest point, v = 0 because the ball just stops momentarily.
Time to top from (7.5): 0 = 20 + (−10)t ⇒ t = 2 s.
Height from (7.6): s = ut + (1/2)at² = 20 × 2 + 0.5 × (−10) × 4 = 40 − 20 = 20 m.
Acceleration is negative because gravity acts downward, opposite to initial motion.
Even while going up, the ball experiences constant downward acceleration.
This causes its speed to reduce to zero at the top.
Use consistent signs: take upward as positive, so gravity is negative.
Q8. Two vehicles travel for the same time t = 10 s. Vehicle A: u = 0, a = 2 m/s². Vehicle B: u = 10 m/s, a = 1 m/s². Who covers more distance? Explain your reasoning fully.
Reason: B already has a head start due to higher initial velocity.
Even though A accelerates faster, the initial speed of B dominates in 10 s.
Always compare both ut and (1/2)at² parts to understand outcomes.
Q9. A car speeds up from 10 m/s to 30 m/s under uniform acceleration of 5 m/s². Find the distance without using time. Verify your answer with another equation.
Answer:
Using (7.7): 2as = v² − u² ⇒ 2 × 5 × s = 900 − 100 = 800 ⇒ s = 80 m.
Now verify using time. From (7.5): t = (v − u)/a = (30 − 10)/5 = 4 s.
Using (7.6): s = ut + (1/2)at² = 10 × 4 + 0.5 × 5 × 16 = 40 + 40 = 80 m.
Both methods give the same distance = 80 m.
Equation (7.7) is very useful when time is not given.
It directly links velocity and displacement.
Always check with a second method if possible.
Q10. A train moves in three stages: (i) starts from rest and accelerates at 1 m/s² for 20 s, (ii) moves at constant speed for 30 s, (iii) decelerates at 2 m/s² to rest. Find the total distance.
Answer:
Stage (i): u = 0, a = 1, t = 20 s. v₁ = u + at = 0 + 1 × 20 = 20 m/s. s₁ = ut + (1/2)at² = 0 + 0.5 × 1 × 400 = 200 m.
Stage (ii): Constant speed v = 20 m/s for 30 s. s₂ = v × t = 20 × 30 = 600 m.
Stage (iii): u = 20 m/s, v = 0, a = −2 m/s². Use (7.7): 2as = v² − u² ⇒ 2 × (−2) × s₃ = 0 − 400 ⇒ s₃ = 100 m.
Total distance = s₁ + s₂ + s₃ = 200 + 600 + 100 = 900 m.
Note the use of different equations for different stages.
Keep signs correct during deceleration.
Break complex motion into simple uniform-acceleration parts.
