for all we know the acceleration is constant. I actually suspect the force is greatest when pushing off the starting block and therefore acceleration is greatest at the start (F = m a), but there is nothing in your statement to help us with a versus t. Therefore assume a is constant.
d = (1/2) a t^2
10 = (1/2) a (2^2)
20 = 4 a
a = 5 m/s^2 indeed
An athlete is training and is required to perform shuttle runs. From the starting position they sprint out to 10m (this takes 2 seconds). I know that a = 5m/s/s but I need to know at what metre within the run would maximum acceleration occur. Appreciate your help.
thanks
2 answers
I think I know what you are looking for, so starting off where Damon left off:
Max acceleration occurs when the velocity=acceleration.
Vi=0m/s^2
Vf=5m/s and
a=5m/s^2
d=?
Solve for d.
Vf^2=Vi^2+2ad
(5m/s)^2=0+2(5m/s^2)d
25=10d
25/10=d
d=2.5m
The sprinters will not keep accelerating indefinitely, so I think this what your teacher/professor had in mind. But a better question would be when will the velocity equal the acceleration.
Best
BTW, I answered this in your earlier post.
Max acceleration occurs when the velocity=acceleration.
Vi=0m/s^2
Vf=5m/s and
a=5m/s^2
d=?
Solve for d.
Vf^2=Vi^2+2ad
(5m/s)^2=0+2(5m/s^2)d
25=10d
25/10=d
d=2.5m
The sprinters will not keep accelerating indefinitely, so I think this what your teacher/professor had in mind. But a better question would be when will the velocity equal the acceleration.
Best
BTW, I answered this in your earlier post.