Question #05757

1 Answer
A08
Feb 10, 2017

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

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