A point initially at rest moves along x -axis. Its acceleration varies with time as a=(6 t+5) in m / s ^2 . If it starts from origin, the distance covered in 2 s is :

Step-by-step solution

Given, a= d v d t =6 t+5 Integrating it, ₀^v d v= ₀^t(6 t+5) d tv= 6 t^2 2 +5 t As v= d s d t , so d s= ( 6 t^2 2 +5 t ) d t Integrating it, ₀^s d s= ₀^t ( 6 t^2 2 +5 t ) d ts= 3 t^3 3 + 5 t^2 2 When aligned t & =2 ~s , s=3 (2)^3 3 + 5(2)^2 2 & =8+10=18 ~m aligned