It is a case of non-uniform acceleration. (Assuming that tt is acceleration is time).
We know that
Acceleration a=(dv(t))/dt=-0.9ta=dv(t)dt=−0.9t
=>dv(t)=-0.9tcdot dt⇒dv(t)=−0.9t⋅dt
Integrating both sides we get
=>v(t)=-0.9int tcdot dt⇒v(t)=−0.9∫t⋅dt
=>v(t)=-0.9(t^2/2+C)⇒v(t)=−0.9(t22+C) .....(1)
where CC is a constant of integration. It is given that the initial velocity at t=0t=0 is 23" ms"^-123 ms−1. Inserting in (1)
v(0)=23=-0.9(0^2/2+C)v(0)=23=−0.9(022+C)
=>-0.9C=23⇒−0.9C=23
=>C=-23/0.9⇒C=−230.9
Expression for velocity becomes
=>v(t)=-0.9(t^2/2-23/0.9)⇒v(t)=−0.9(t22−230.9)
=>v(t)=-0.45t^2+23⇒v(t)=−0.45t2+23 ......(2)
To find the velocity after 6s6s. From (2) we get
v(6)=-0.45xx6^2+23v(6)=−0.45×62+23
v(6)=6.8" ms"^-1v(6)=6.8 ms−1
