Question

A particle starts from P and reaches at Q. Its equation of motion is given by s=t^3/3-2t^2.Find
a)velocity
b)time taken by it to reach at Q​​

Answer 1

Answer:

Position of the particle x=2t

2

−t

3

Velocity v=

dt

dx

=4t−3t

2

Acceleration a=

dt

dv

=4−6t

So, a( at t=2 s)=4−6×2=−8 m/s

2

Since, acceleration is depending on the time. SO, acceleration is not constant.