Question

The position of the particle is given by x=6t2-2t+3. Find (a) Its initial position (b) instantaneous velocity at 3s (c) instantaneous acceleration at t=2s (d) average velocity between t=2s and t=3s

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.