Question

The position of a particle moving along x direction varies as x=2t^3-24t+6. Then find the "distance" travelled by the body during t=1 sec to t=4 sec.

Answer 1

Hints : First find dx/dt and then check when the velocity becomes zero, then find the displacement during that time and then again during the remaining time, add them to get your answer.

Solution : (dx/dt) = v = 6t² - 24

Here, v is zero at 2s.

x from 1s to 2s (during 1s)
= x at 2s - x at 1s
= -26 - (-16) = -10 m

x during 2s
= 2×8 - 24×2 + 6
= 16 - 48 + 6
= -26 m

.°. Distance travelled during 4s = 10+26 = 36m.

P. S. : Hope you understand. Please comment for any doubts or if the answer is incorrect.