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

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

Step-by-step 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 = (v_(f)^(2) - v_(0)^(2))/(2a) = (-(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!