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:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex] v_{f}[/tex]: is the final velocity = 0 (when it stops)
[tex] v_{0}[/tex]: 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:
[tex] v_{f} = v_{0} + at [/tex]
[tex] 0 = 30 m/s + (-4t m/s^{2})t [/tex]
[tex]0 = 30 m/s - 4t^{2} m/s^{3}[/tex]
[tex]t = 2.74 s[/tex]
Now, the acceleration is:
[tex] a = -4t = -10.96 m/s^{2} [/tex]
Hence, the distance is:
[tex] d = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{-(30 m/s)^{2}}{2*(-10.96 m/s^{2})} = 41.06 m [/tex]
Therefore, the distance traveled by the particle before it stops is 41.06 m.
I hope it helps you!
