Question #05757

1 Answer
Feb 10, 2017

Answer:

#6.8" ms"^-1#

Explanation:

It is a case of non-uniform acceleration. (Assuming that #t# is acceleration is time).
We know that
Acceleration #a=(dv(t))/dt=-0.9t#
#=>dv(t)=-0.9tcdot dt#

Integrating both sides we get
#=>v(t)=-0.9int tcdot dt#
#=>v(t)=-0.9(t^2/2+C)# .....(1)
where #C# is a constant of integration. It is given that the initial velocity at #t=0# is #23" ms"^-1#. Inserting in (1)
#v(0)=23=-0.9(0^2/2+C)#
#=>-0.9C=23#
#=>C=-23/0.9#
Expression for velocity becomes
#=>v(t)=-0.9(t^2/2-23/0.9)#
#=>v(t)=-0.45t^2+23# ......(2)

To find the velocity after #6s#. From (2) we get
#v(6)=-0.45xx6^2+23#
#v(6)=6.8" ms"^-1#