Question

A particle moves along a straight line in such a way that its acceleration is increasing at the rate of $2m/s^3$. Its initial acceleration and velocity were zero. Then, the distance which it will covers in $3^{st}$ second is:-

• A. 9 m
• B. $\frac{12}{5}m$
• C. $\frac{17}{5}m$
• D. $\frac{19}{4}m$

Answer 1

The correct option is A 9 m

$\frac{da}{at}=2$

$a=2t+c$

at $t=0,a=0$ so $c=0$

$a=2t$

using: $v=\int adt$ & at $t=0,v=0$

$v=t^2$

using: $s=\int vdt$

$s={\int }_0^3{t}^{2}dt$

$s=\frac{3^3}{3}$

$s=9m$