A particle moves in a straight line and its distance S(m) from a point is given by S=45t+11t-t^3. Find. expression of velocity in terms of t,the expression for acceleration in terms of t and the velocity and acceleration when t= 3 seconds?

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Answer 1 · 2017-08-30T05:17:02

See the explanation below

The distance is

#s=45t+11t-t^3#

The velocity is the derivative of the distance

#v(t)=(ds)/(dt)=45+11-3t^2=56-3t^2#

The acceleration is the derivative of the velocity

#a(t)=(dv)/(dt)=-6t#

Therefore, when #t=3s#

#v(3)=56-3*9=56-27=29ms^-1#

and

#a(3)=-6*3=-18ms^-2#