Questions and Answers (Subjective)

Q1. Define displacement and distance. How are they different?

Ans:

• Distance is the total length of the path traveled by an object (scalar).
• Displacement is the shortest path between initial and final positions (vector).
• Displacement can be zero or negative, but distance is always positive.

Q2. Define speed and velocity. Which can be zero: average speed or average velocity?

Ans:

• Speed = rate of distance covered (scalar)
• Velocity = rate of displacement (vector)
• Average velocity can be zero if displacement is zero, but average speed cannot be zero unless distance is zero.

Q3. Define instantaneous velocity. How is it obtained from position-time graph?

Ans: It is the velocity at a specific instant: $$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$ It is equal to the slope of tangent to the x–t graph.

Q4. A particle travels half of its journey at 60 km/h and remaining half at 40 km/h. Find average speed.

Ans: Let each half be of distance $d$: Time for first half = $d/60$, second = $d/40$ Total time = $\frac{d}{60} + \frac{d}{40} = \frac{5d}{120} = \frac{d}{24}$

Average speed = $\frac{2d}{(d/24)} = 48$ km/h

Q5. Derive the first equation of motion: $v = u + at$ using calculus.

Ans: From definition: $$a = \frac{dv}{dt} ⇒ dv = a dt ⇒ \int_u^v dv = \int_0^t a dt$$ $$v - u = at ⇒ v = u + at$$

Q6. Derive $s = ut + \frac{1}{2}at^2$ using calculus.

Ans: $v = \frac{dx}{dt} = u + at$ Integrate: $$\int dx = \int (u + at) dt ⇒ x = ut + \frac{1}{2}at^2 + C$$ Assuming $x=0$ at $t=0$, $C = 0$ ⇒ $x = ut + \frac{1}{2}at^2$

Q7. Derive $v^2 = u^2 + 2as$ using algebra.

Ans: From: $$s = ut + \frac{1}{2}at^2\ \text{and}\ v = u + at$$ Eliminate $t$: $$t = \frac{v - u}{a} ⇒ s = \frac{u(v - u)}{a} + \frac{1}{2}a\left(\frac{v - u}{a}\right)^2$$ Simplifying: $$v^2 = u^2 + 2as$$

Q8. Explain the meaning of area under v–t and a–t graphs.

Ans:

• Area under v–t graph = displacement
• Area under a–t graph = change in velocity

**Q9. Draw x–t graph for:

1. Uniform motion
2. Uniform acceleration**

Ans:

• Uniform motion: Straight line with constant slope
• Uniform acceleration: Parabolic curve

Q10. A car moves with 10 m/s for 5s, then accelerates at 2 m/s² for 5s. Find total displacement.

Ans:

• First phase: $s_1 = v × t = 10 × 5 = 50$
• Second phase: $u = 10$, $t = 5$, $a = 2$ $$s_2 = ut + \frac{1}{2}at^2 = 10×5 + 0.5×2×25 = 50 + 25 = 75$$ Total $s = 125$ m

Q11. What is relative velocity? Derive expression for two bodies moving in same direction.

Ans: Relative velocity of A with respect to B: $$v_{AB} = v_A - v_B$$ If both move in same direction, subtract their speeds.

Q12. Two objects move toward each other at 30 m/s and 50 m/s. Find their relative velocity.

Ans: Opposite directions: $$v_{rel} = v_A + v_B = 30 + 50 = 80\ \text{m/s}$$

Q13. Define acceleration. What does negative acceleration indicate?

Ans: Acceleration is the rate of change of velocity: $$a = \frac{dv}{dt}$$ Negative acceleration means deceleration — object is slowing down.

Q14. Find acceleration from $v = 5t^2 + 3t$.

Ans: Acceleration is: $$a = \frac{dv}{dt} = 10t + 3$$

Q15. A train starts from rest and gains speed uniformly at 0.5 m/s² for 40s. Find distance covered.

Ans: Given: $u = 0$, $a = 0.5$, $t = 40$ $$s = ut + \frac{1}{2}at^2 = 0 + 0.5×1600×0.5 = 400\ \text{m}$$

Q16. Define jerk. How is it related to acceleration?

Ans: Jerk is the rate of change of acceleration: $$\text{Jerk} = \frac{da}{dt}$$ It measures how quickly acceleration changes.

Q17. State the conditions under which displacement equals distance.

Ans: When an object moves in a straight line and does not change direction, displacement = distance.

Q18. Can displacement be zero when distance is not zero? Explain with example.

Ans: Yes. If a person walks 5 m east and returns 5 m west:

• Distance = 10 m
• Displacement = 0 m (final position = initial)

Q19. Explain how calculus is used to define motion quantities.

Ans:

• Velocity = derivative of position: $v = dx/dt$
• Acceleration = derivative of velocity: $a = dv/dt$
• Displacement = integral of velocity
• Change in velocity = integral of acceleration

Q20. A bullet travels 100 m in 0.5 s. Find its average speed.

Ans: $$\text{Average speed} = \frac{\text{Distance}}{\text{Time}} = \frac{100}{0.5} = 200\ \text{m/s}$$