Question

8) For the displacement
s= 3t^3+4t^2+9t+2, find

i) velocity at t = 2sec
ii) average velocity 1 to 3 sec.

iii) acceleration at t = 2 sec
iv) average acceleration 1 to 3 sec

v) time and acceleration when body is at rest.​

Answer 1

Answer:

s=t

3

−6t

2

+3t=4

v=

dt

ds

=3t

2

−12t+3

a=

dt

dv

=6t−12

So, acceleration will be zero at

6t−12=0

t=2sec

Then velocity at t = 3

v=3(2)

2

−12(2)+3

= 12-24+3

v=−9m/s