Question

If the velocity of a particle varies as v = 4t, find its displacement in first three seconds.

• A. 9 m
  
• B. 18 m
• C. 2 m
• D. 12 m

Answer 1

The correct option is B 18 m

$v=\frac{ds}{dt}$
$ds=vdt=4tdt$
$s={\int }_0^{3}ds={\int }_0^{3}4tdt=[\frac{4.t^2}{2}{]}_0^3=[2t^2{]}_0^3$
$=2(9-0)$
$=18m$