Given: The velocity function of a moving particle as
[tex]V(t)=e^{2sint}-1[/tex]
To Determine: The time at which the acceleration equals to zero
Solution
Note that at time t=0, the particle is at origin, so
[tex]\begin{gathered} V(0)=e^{2sin0}-1 \\ V(0)=e^{2\times0}-1 \\ V(0)=e^0-1 \\ V(0)=1-1 \\ V(0)=0 \end{gathered}[/tex]
Determine the acceleration function
The acceleration of a particle is the rate of change of velocity or the derivative of the velocity function. Therefore,
[tex]A(t)=\frac{dV(t)}{dt},or,A(t)=V^{\prime}(t)[/tex][tex]\begin{gathered} V(t)=e^{2sint}-1 \\ let:u=2sint:\frac{du}{dt}=2cost \\ V(t)=e^u \\ \frac{dV(t)}{du}=e^u=e^{2sint} \\ A(t)=\frac{dV(t)}{dt}=\frac{dV}{du}\times\frac{du}{dt} \\ A(t)=e^{2sint}\times2cost \\ A(t)=2e^{2sint}cost \end{gathered}[/tex]
When the acceleration is equal to zero, then we have
[tex]\begin{gathered} A(t)=0 \\ 2e^{2sint}cost=0 \end{gathered}[/tex]
Let us plot the graph of the acceleration function
The time for given interval for which the acceleration is zero are
[tex]t=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2},\frac{9\pi}{2}[/tex]
ence, thre first time the acceleration is zero is π/2 or 1.571
