A crate is initially at rest on a rough, horizontal floor. The crate moves in a straight line as it is pushed resulting in the crate speeding up at a constant rate of 3.89 m/s2. After a while, the crate is no longer being pushed and the crate slows down at a constant rate of 5.48 m/s2. The crate comes to rest a total distance 51.6 m away from its starting position. For how long (amount of time) was the crate pushed? Model this problem as having two constant accelerations as described above. Write your answer to three significant figures.
To solve this problem, we need to use the equations of motion and the given information.
Let's denote the time the crate was pushed as t1 and the time it was slowed down as t2.
We are given that the crate speeds up at a constant rate of 3.89 m/s², so we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity (0 m/s as the crate starts at rest), a is the acceleration, and t is the time.
Using this equation, we can find the final velocity when the crate stops being pushed:
v1 = u + a1 * t1
0 m/s = 0 m/s + (3.89 m/s²) * t1
Solving for t1:
t1 = -u / a1
t1 = 0 / 3.89 m/s²
t1 = 0 s
This tells us that the crate stops being pushed instantaneously, so t1 = 0 s.
Now, we know that the crate slows down at a constant rate of 5.48 m/s². Using the same equation, we can find the time it takes for the crate to come to rest:
v2 = u + a2 * t2
0 m/s = v1 + (-5.48 m/s²) * t2
0 m/s = 3.89 m/s² * t2
t2 = -v1 / a2
t2 = -3.89 m/s² / -5.48 m/s²
t2 ≈ 0.71 s
So, the crate was pushed for t1 + t2 = 0 s + 0.71 s = 0.71 s.
Therefore, the crate was pushed for approximately 0.71 seconds.