Question

Body A is moving with uniform velocity of 8 m/s along a straight line. When it crosses B, B starts from rest and moves with uniform acceleration of 4 m/s? along the same straight line. The time after which the two bodies meet will be:

• A. 2 s
• B. 4 s
• C. 6 s
• D. 8 s

Answer 1

Answer: (B)

4 s

Answer 2

Let the initial position where Body A crosses Body B be the origin (x=0) at time t=0.

For Body A (uniform velocity):

• Initial position $x_{A,0} = 0$
• Velocity $v_A = 8 \, \text{m/s}$
• Position at time t: $x_A(t) = x_{A,0} + v_A t = 0 + 8t = 8t$

For Body B (starts from rest with uniform acceleration):

• Initial position $x_{B,0} = 0$
• Initial velocity $u_B = 0 \, \text{m/s}$
• Acceleration $a_B = 4 \, \text{m/s}^2$
• Position at time t: $x_B(t) = x_{B,0} + u_B t + \frac{1}{2} a_B t^2 = 0 + 0 \cdot t + \frac{1}{2}(4)t^2 = 2t^2$

The two bodies meet when their positions are equal, i.e., $x_A(t) = x_B(t)$.
$8t = 2t^2$
$2t^2 - 8t = 0$
$2t(t - 4) = 0$

This equation yields two solutions for t:

1. $t = 0 \, \text{s}$: This is the initial moment when Body A crosses Body B.
2. $t = 4 \, \text{s}$: This is the time after which the two bodies meet again.

The question asks for the time after which the two bodies meet, which refers to the second instance of meeting.
Therefore, the time after which the two bodies meet will be 4 seconds.