Question

A particle moves along the r-axis so that its velocity vat any given time t, for 0 St $ 16, is given byW(0) =e»-1. At time r=0, the particle is at the origin. At what time does the particle's accelerationequal zero for the first time?Round to the nearest thousandth or write as a fraction

Answer 1

Given: The velocity function of a moving particle as

`V(t)=e^(2sint)-1`

To Determine: The time at which the acceleration equals to zero

Solution

Note that at time t=0, the particle is at origin, so

`\begin{gathered} V(0)=e^(2sin0)-1 \\ V(0)=e^(2*0)-1 \\ V(0)=e^0-1 \\ V(0)=1-1 \\ V(0)=0 \end{gathered}`

Determine the acceleration function

The acceleration of a particle is the rate of change of velocity or the derivative of the velocity function. Therefore,

`A(t)=(dV(t))/(dt),or,A(t)=V^(\prime)(t)`
`\begin{gathered} V(t)=e^(2sint)-1 \\ let:u=2sint:(du)/(dt)=2cost \\ V(t)=e^u \\ (dV(t))/(du)=e^u=e^(2sint) \\ A(t)=(dV(t))/(dt)=(dV)/(du)*(du)/(dt) \\ A(t)=e^(2sint)*2cost \\ A(t)=2e^(2sint)cost \end{gathered}`

When the acceleration is equal to zero, then we have

`\begin{gathered} A(t)=0 \\ 2e^(2sint)cost=0 \end{gathered}`

Let us plot the graph of the acceleration function

The time for given interval for which the acceleration is zero are

`t=(\pi)/(2),(3\pi)/(2),(5\pi)/(2),(7\pi)/(2),(9\pi)/(2)`

ence, thre first time the acceleration is zero is π/2 or 1.571