An object moves in a straight line with deceleration whose magnitude varies with velocity as 3 v^ 2 / 3 . If at an initial point, the velocity is 8 ~m / s , then the distance travelled by the object before it stops is

Step-by-step solution

Given that, a= deceleration =-3 v^ 2 / 3 ~m / s ^2 At initial point (t=0) velocity =8 ~m / s Distance travelled by the object before it stop is We know that, aligned a & = d v d t a= d v d s d s d t a & =v d v / d s a d s & =v d v -3 v^ 2 / 3 d s & =v d s d d s & =- 1 3 v^ 1 / 3 d v aligned Integrate both sides, we get aligned & d s= -1 3 v^ 1 / 3 d v & s= -1 4 v^ 4 / 3 +C & At t=0, u=8 ~m / s , s=0 & C= 1 4 (8)^ 4 / 3 = 16 4 =4 & aligned So, Therefore, from Eq. (i) s= -1 4 v^ 4 / 3 +4 Body stops, when v=0 . So, s=