You are trying to get to class on time using the UCF Shuttle. You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. In 40.9 m you will reach a barrier and you must catch the shuttle before that point. The shuttle has a constant acceleration of 4.5 m/s2. What is the minimum velocity you have to run at to catch the bus before it reaches the barrier

20.1 m/s

Explanation:

Since You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. And In 40.9 m you will reach a barrier and you must catch the shuttle before that point.

Given that the shuttle has a constant acceleration of 4.5 m/s2. 

The total distance to cover is:

Total distance = 40.9 + 3.9 = 44.8 m

Assuming you are starting from rest. Then initial velocity U = 0

Using the 3rd equation of motion to calculate the minimum velocity.

V^2 = U^2 + 2as

V^2 = 0 + 2 × 4.5 × 44.8

V^2 = 403.2

V = sqrt (403.2)

V = 20.1 m/s

Therefore, the minimum velocity you have to run at to catch the bus before it reaches the barrier is 20.1 m/s

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