A car starts from rest and moves a distance of 5m in t seconds ,where s=1/6t³+1/4t².find the acceleration when t=0,t=2?

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Answer 1 · 2017-11-29T13:20:54

See the explanation below

I assumed that the #5# is #s#

Start by calculating the velocity, which is the derivative of the position

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

#v(t)=s'(t)=3/6t^2+2/4t=1/2t^2+1/2t#

The acceleration is the derivative of the velocity

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

When #t=0#

The acceleration is #a(0)=1/2ms^-2#

When #t=2#

The acceleration is #a(2)=5/2ms^-2#