A car starts from rest and moves a distance of 5m in t seconds,where s=1/6t^3+1/4t^2.what is its acceleration when t=0,t=2?

2 Answers
Jan 14, 2018

See the explanation below

Explanation:

I assume that #5# means #s#

The acceleration is the second derivative of the distance, the first derivative is the velocity.

#s(t)=1/6t^3+1/4t^2#

The velocity is

#v(t)=s'(t)=1/2t^2+1/2t#

The acceleration is

#a(t)=v'(t)=t+1/2#

Therefore,

When #t=0#, #=>#, #a(0)=1/2ms^-2#

When #t=2#, #=>#, #a(2)=5/2ms^-2#

I hope that this helps!!!!

Jan 14, 2018

Acceleration is #0.5 m/(s^2)# at t=0
Acceleration is #2.5 m/(s^2)# at t=2

Explanation:

Differentiate the given equation twice and you will get,acceleration of the car to be,

#a = t + (1/2) #
Now put the given time values it comes out to be 0.5 # m/(s^2)# at t=0 and #2.5 m/(s^2)# at t=2