Question

If the particle's velocity is governed by: v = 2s +1 where s is in [ft] and v is in [ft/s], Find the particle's acceleration at s = 3 ft. Find how long it takes the particle to reach 100 ft if its initial position was 0 ft.

Answer 1

The particle takes 3.41 sec to reach 100 ft.

Step-by-step explanation:

Given that,

Velocity v= 2s+1

Distance s = 100 ft

Acceleration :

The acceleration is the first derivative of the velocity of the particle.

`a =(dv)/(dt)`

But,
`v=(ds)/(dt)=2s+1`

`(ds)/(2s+1)=dt`

Multiply by 2 in both side

`(2ds)/(2s+1)=2dt`

On integrating both side

`\int{(2ds)/(2s+1)}=\int{2dt}`

`log(2s+1)=2t+C`

`2s+1=e^(2t+C)`

`v=e^(2t+C)`

On differentiating w.r.to t

`(dv)/(dt)=2e^(2t+C)`

The acceleration at s = 3

`a=2(2s+1)`

`a=2(2*3+1)`

`a=14\ ft/s^2`

(II). We need to calculate the time

Using equation of motion

`s=ut+(1)/(2)at^2`....(II)

We need to calculate the initial velocity

The particle's velocity is

v= 2s+1

Put the value of s in the equation

`u=2*0+1`

`u=1\ ft/s`

Now, Put the value in the equation (II)

`100=1* t+(1)/(2)*14* t^2`

`7t^2+t-100=0`

`t=3.71,-3.81`

t can not be negative.

`t = 3.71\ sec`

Hence, The particle takes 3.41 sec to reach 100 ft.