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

Answer:

The particle takes 3.41 sec to reach 100 ft.

Explanation:

Given that,

Velocity v= 2s+1

Distance s = 100 ft

Acceleration :

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

$a =\dfrac{dv}{dt}$

But, $v=\dfrac{ds}{dt}=2s+1$

$\dfrac{ds}{2s+1}=dt$

Multiply by 2 in both side

$\dfrac{2ds}{2s+1}=2dt$

On integrating both side

$\int{\dfrac{2ds}{2s+1}}=\int{2dt}$

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

$2s+1=e^{2t+C}$

$v=e^{2t+C}$

On differentiating w.r.to t

$\dfrac{dv}{dt}=2e^{2t+C}$

The acceleration at s = 3

$a=2(2s+1)$

$a=2(2\times3+1)$

$a=14\ ft/s^2$

(II). We need to calculate the time

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$....(II)

We need to calculate the initial velocity

The particle's velocity is

v= 2s+1

Put the value of s in the equation

$u=2\times0+1$

$u=1\ ft/s$

Now, Put the value in the equation (II)

$100=1\times t+\dfrac{1}{2}\times14\times t^2$

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

$t=3.71,-3.81$

t can not be negative.

$t = 3.71\ sec$

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