Question

The position of a particle moving along x-axis is given by x = 10t -$2t^2$. Then the time (t) at which it will momently come to rest is

• A. 0
• B. 2.5 s
• C. 5 s
• D. 10 s

Answer 1

The correct option is B 2.5 s

Given that,

$x=10t-2t^2$

$so,$

$v=\frac{dx}{dt}=10-4t$

The particle will come to rest if v = 0

$10-4t=0$

$t=2.5s$