Numerical Problems Based on Motion with Uniform Acceleration for Class 11 Physics

Initial velocity, $u = 0$ (starts from rest).
Final velocity, $v = 180\text{ kmh}^{-1} = 180 \times \frac{5}{18}\text{ ms}^{-1} = 50\text{ ms}^{-1}$.
Time, $t = 25\text{ s}$.

First, find acceleration ($a$) using $v = u + at$:

Now, calculate distance ($s$) using $s = ut + \frac{1}{2}at^2$:

Initial velocity, $u = 16\text{ ms}^{-1}$.
Final velocity, $v = 0$ (comes to rest).
Distance, $s = 0.4\text{ m}$.

First, find the retardation ($a$) using $v^2 – u^2 = 2as$:

Now, find time ($t$) using $v = u + at$:

Initial velocity, $u = 72\text{ kmh}^{-1} = 72 \times \frac{5}{18}\text{ ms}^{-1} = 20\text{ ms}^{-1}$.
Final velocity, $v = 0$.
Distance, $s = 100\text{ m}$.

Retardation: Use $v^2 – u^2 = 2as$

Therefore, retardation is $2\text{ ms}^{-2}$.

Time: Use $v = u + at$

Initial velocity, $u = 36\text{ kmh}^{-1} = 36 \times \frac{5}{18}\text{ ms}^{-1} = 10\text{ ms}^{-1}$.
Final velocity, $v = 0$.
The car must stop 5 m before the child, so stopping distance $s = 55 – 5 = 50\text{ m}$.

Retardation: Use $v^2 – u^2 = 2as$

Retardation is $1\text{ ms}^{-2}$.

Time: Use $v = u + at$

Initial velocity, $u = 30\text{ kmh}^{-1} = 30 \times \frac{5}{18} = \frac{25}{3} \approx 8.33\text{ ms}^{-1}$.
Reaction time, $t_r = 0.6\text{ s}$.
Deceleration, $a = -5\text{ ms}^{-2}$.

Total distance = Distance during reaction time + Braking distance.

1. Distance during reaction time ($s_1$):
During this time, the car travels at constant speed $u$.

2. Braking distance ($s_2$):
Using $v^2 – u^2 = 2as$ with $v = 0$:

Total distance:

This motion happens in three stages:

Stage 1 (Acceleration): Starts from rest $u_1=0$, final velocity $v_1=8\text{ ms}^{-1}$, time $t_1=10\text{ s}$.

Stage 3 (Retardation): Initial velocity $u_3 = 8\text{ ms}^{-1}$, final velocity $v_3=0$, distance $s_3 = 64\text{ m}$.

Time for stage 3: $v_3 = u_3 + a_3t_3 \implies 0 = 8 – 0.5t_3 \implies t_3 = 16\text{ s}$.

Stage 2 (Constant Velocity):
Total distance $S = 584\text{ m}$. Distance for stage 2 is $s_2 = S – (s_1 + s_3)$.

Since velocity is constant at $8\text{ ms}^{-1}$:

Total Time:

Let’s calculate the stopping distance for both trains individually.

Train A:
Initial velocity, $u_A = 72\text{ kmh}^{-1} = 72 \times \frac{5}{18} = 20\text{ ms}^{-1}$.
Final velocity $v_A = 0$, retardation $a_A = -1.0\text{ ms}^{-2}$.

Train B:
Initial velocity, $u_B = 90\text{ kmh}^{-1} = 90 \times \frac{5}{18} = 25\text{ ms}^{-1}$.
Final velocity $v_B = 0$, retardation $a_B = -1.0\text{ ms}^{-2}$.

Total stopping distance:

The initial distance between them is $1.0\text{ km} = 1000\text{ m}$. Since the total distance they require to stop ($512.5\text{ m}$) is less than the distance between them ($1000\text{ m}$), they will not collide.