Question

A particle is fallingdown into a medium with an initial velocity of 30m/s. If the acceleration of the particle is =(−4t)/m/s^2 , where t is in seconds, determine the distance traveled before the particle stops.

Answer 1

Answer:

The distance traveled by the particle before it stops is 41.06 m.

Explanation:

We can find the distance traveled by the particle using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

Where:

$v_{f}$: is the final velocity = 0 (when it stops)

$v_{0}$: is the initial velocity = 30 m/s

a: is the acceleration = -4t m/s²

t: is the time

d: is the distance

First, we need to calculate the time:

$v_{f} = v_{0} + at$

$0 = 30 m/s + (-4t m/s^{2})t$

$0 = 30 m/s - 4t^{2} m/s^{3}$    

$t = 2.74 s$

Now, the acceleration is:

$a = -4t = -10.96 m/s^{2}$                    

Hence, the distance is:          

$d = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{-(30 m/s)^{2}}{2*(-10.96 m/s^{2})} = 41.06 m$      

Therefore, the distance traveled by the particle before it stops is 41.06 m.

                     

I hope it helps you!