Question

A cyclist racing in the keiran is riding at the top of the track at 5m/s. Then he sprints downhill in the sprinting to the finish in 1.75s. His acceleration was 9m/s^2. What was his ending velocity at the finish line?

Answer 1

The cyclist is moving by uniformly accelerated motion, with an initial velocity of
`v_i=5~m/s` and an acceleration of
`a=9~m/s^2`.
The acceleration is given by

`a= (v_f-v_i)/(\Delta t)`
where
`v_f` is the final velocity and
`\Delta t` is the time between the end and the beginning of the motion, and in our case it is 1.75 s. Therefore, from this relationship we can find the final velocity:

`v_f=v_i + a \Delta t=5~m/s+9~m/s^2 \cdot 1.75~s=20.75~m/s`