Question

part a. A nerf dart is launched out of a pvc pipe, and while it is accelerating it goes from rest to a velocity of 30.00 m/s. The acceleration of the dart is 330 m/s^2. How long is the dart being accelerated?part b. How far does the dart travel being accelerated?

Answer 1

Part A. We are given that a dart travels from rest to a velocity of 30 m/s with an acceleration of 330 m/s^2 to determine the time we will use the following equation of motion:

`v_f=v_0+at`

Where:

`\begin{gathered} v_f,v_0=\text{ final and initial velocities} \\ a=\text{ acceleration } \\ t=\text{ time} \end{gathered}`

Since the dart is launched from rest this means that the initial velocity is zero, therefore:

`v_f=at`

Now, we divide both sides by "a":

`(v_f)/(a)=t`

Now, we plug in the values:

`(30(m)/(s))/(330(m)/(s^2))=t`

Solving the operation:

`0.09s=t`

Part B. Now, we are asked to determine the distance. To do that we will use the following equation of motion:

`2ad=v_f^2-v_0^2`

Now, we divide both sides by "2a":

`d=\frac{v_f^2-v_0{}^2}{2a}`

Now, we plug in the values:

`d=((30(m)/(s))^2)/(2(330(m)/(s^2)))`

Solving the operations we get:

`d=1.36m`

Therefore, the distance is 1.36 meters.