Question

x= 2t^3 + t^2+ t+2
find the acceleration at 2 sec​

Answer 1

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.