Moby then goes up the other side of the halfpipe and stops. If that trip also took five seconds, what was Moby's acceleration?
To find Moby's acceleration, we need to use the formula:
Acceleration (a) = Change in Velocity (Δv) / Time (Δt)
Given that Moby goes up the other side of the halfpipe and stops, this indicates a change in velocity (Δv) from his initial velocity (Vi) to a final velocity (Vf) of zero.
Time is given as 5 seconds for both trips.
Therefore, we can calculate the acceleration as follows:
Acceleration (a) = (Vf - Vi) / Δt
Since Moby's initial velocity (Vi) is not given, we assume it to be zero, as he starts from rest.
Acceleration (a) = (0 - 0) / 5
Acceleration (a) = 0 / 5
Acceleration (a) = 0 m/s²
So, Moby's acceleration is 0 m/s².
To calculate Moby's acceleration, we need to use the formula for acceleration: acceleration = (final velocity - initial velocity) / time.
Given that Moby goes up the other side of the halfpipe and stops, we know that his final velocity is 0 m/s because he comes to a stop. We also know that the total time for both trips is 5 seconds.
Now, we need to determine Moby's initial velocity. Since he starts from rest at the bottom of the halfpipe, his initial velocity is also 0 m/s.
Plugging these values into the formula: acceleration = (0 m/s - 0 m/s) / 5 s.
As the numerator is 0, the acceleration will also be 0 regardless of the value in the denominator. Therefore, Moby's acceleration is 0 m/s².