a cyclist bikes at a speed of 6 m/s after rolling down a hill in 3 s. Of the cyclist has a constant acceleration of 2 m/s during the descent, what was the velocity of the cyclist at the top of the hill?
since a = 2m/s^2 , note you have a typo there.
v = 2t + c
when t = 3, v = 6
6 = 2(3) + c
c = 0
v = 2t
so at top of hill, t = 0
then v = 0
To find the velocity of the cyclist at the top of the hill, we need to use the kinematic equation of motion:
vf = vi + at
Where:
- vf represents the final velocity of the cyclist at the bottom of the hill
- vi represents the initial velocity of the cyclist at the top of the hill
- a represents the acceleration during the descent
- t represents the time taken to travel down the hill
Given:
- vf = 6 m/s (final velocity at the bottom of the hill)
- a = 2 m/s² (acceleration during the descent)
- t = 3 s (time taken to travel down the hill)
First, let's rearrange the kinematic equation to solve for vi:
vi = vf - at
Substituting the given values into the equation:
vi = 6 m/s - (2 m/s²)(3 s)
Simplifying the equation:
vi = 6 m/s - 6 m/s
Finally, solving for vi:
vi = 0 m/s
Therefore, the velocity of the cyclist at the top of the hill is 0 m/s.