The first phase is easy if you have constant acceleration because the average speed until the stop is Vi/2 = 5 m/s
t to stop = 148.3 / 5 = 29.7 seconds
a = change in velocity/change in time = -10/29.7 = -.337 m/s^2
phase 2 displacement = 37.7 - 148.3
= -110.6 meters
-110.6 = (1/2)(-.337)t^2
solve for t
add that t to 29.7 seconds to get the total time
An object has a initial velocity of 10 m/s in the positive x direction. Under a constant acceleration it stops after undergoing a displacement of 148.3 m in the positive x direction. Because it is still under the same acceleration it then travels back in the negative in the x direction and ends up with a displacement of 36.7 m in the positive x direction relative to where it started. What is the time needed for the entire trip in seconds? (From the start until it arrives back where the displacement is 36.7 m in the positive x direction.)
Hint: It's easier to do this in two parts. First find the time until it stops, and then, in the second part, find the time it takes to get from the displacement of 148.3 m until 36.7 m in the positive x direction. Then sum the two times.
1 answer
1 answer