Two forces act on an object. One force is 3 N to the north and the other force is 9N to the south. What is the net force acting on the object? *

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Ответ:

axleon11

05.12.2020

Physics

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!

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