Question

Two particles A and B at rest initially, start to move simultaneously along the same straight line, A with constant velocity \(5\,ms^{-1}\) and B with constant acceleration \(2\,ms^{-2}\). Then the time after which B overtakes A is}

• A. 5 s
• B. 10 s
• C. 15 s
• D. 20 s
• E. 25 s

Hint:

For overtaking problems, equate the displacements of both objects at the same time.

Answer 1

The Correct Option is A

Particle A moves with constant velocity: \[ x_A=vt=5t \] Particle B starts from rest with acceleration \(2\,ms^{-2}\): \[ x_B=\frac12 at^2=\frac12(2)t^2=t^2 \] B overtakes A when: \[ x_A=x_B \] So, \[ 5t=t^2 \] \[ t(t-5)=0 \] Ignoring \(t=0\), we get: \[ t=5\text{ s} \]
Hence, the correct answer is: \[ \boxed{(A)\ 5\text{ s}} \]