Question

On a ride called the Detonator at Worlds of Fun in Kansas City, passengers accelerate straight downward from rest to 45 mi/h in 2.2 seconds.What is the average acceleration of the passengers on this ride?

Answer 1

Explanation:

It is given that,

Initial velocity of passengers, u =0

Final velocity of passengers, v = 45 mi/h = 20.11 m/s

Time taken, t = 2.2 s

We need to find the acceleration of the passengers on this ride. It can be calculated using the formula of acceleration as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{20.11}{2.2}$

$a=9.14\ m/s^2$

So, the acceleration of the passengers on this ride is $9.14\ m/s^2$. Hence, this is the required solution.