Question

Two objects, P and Q, travelling in the same direction starts from rest. While the object P starts at time $t=0$ and the object Q starts later at $t=30$ min. The object P has an acceleration of $40 \text{ km/h}^2$. To catch P at a distance of $20$ km, the acceleration of Q should be}

• A. $40 \text{ km/h}^2$
• B. $80 \text{ km/h}^2$
• C. $100 \text{ km/h}^2$
• D. $120 \text{ km/h}^2$
• E. $160 \text{ km/h}^2$

Hint:

When objects start at different times, always adjust the effective time before applying motion equations.

Answer 1

Answer: (E)

Concept:
For motion starting from rest: \[ s = \frac{1}{2}at^2 \]

Step 1: Find time taken by P to reach 20 km.
\[ 20 = \frac{1}{2} \cdot 40 \cdot t^2 \] \[ 20 = 20t^2 \Rightarrow t^2 = 1 \Rightarrow t = 1 \text{ hr} \]

Step 2: Time available for Q.
Q starts $30$ min later = $0.5$ hr delay: \[ t_Q = 1 - 0.5 = 0.5 \text{ hr} \]

Step 3: Apply distance formula for Q.
\[ 20 = \frac{1}{2} a_Q (0.5)^2 \] \[ 20 = \frac{1}{2} a_Q \cdot 0.25 = 0.125 a_Q \]

Step 4: Solve for $a_Q$.
\[ a_Q = \frac{20}{0.125} = 160 \text{ km/h}^2 \]