Kinematics Question 391
Question: A body A begins to move with initial velocity 2 m/sec and continue to moves at a constant acceleration $ a.\Delta t=10 $ seconds after the body A begins to move a body B departs from the same point with an initial velocity 12 m/sec and moves with the same acceleration a. What is the maximum acceleration a at which the body B can overtake A?
Options:
A) $ 1m/s^{2} $
B) $ 2m/s^{2} $
C) $ 1/2m/s^{2} $
D) $ 3m/s^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ a=\frac{v_0-v’_0}{\Delta t} $
On substituting $ v _{0}2\text{ m/s, v }’\text{ }’ _{0}=12\text{ m/s and }\Delta \text{t=10 sec,} $
We get, $ a=1m/s^{2} $
