Question

On an amusement park ride passengers accelerate straight downward from rest to 22.9 m/s in 2.2 s.

What is the average acceleration of the passengers on this ride?

Answer 1

Given,

Initial Velocity (u) = 0 m/s

Final Velocity (v) = 25.9 m/s

Time (t) = 2.5 sec

avg. acceleration = ?

avg. acceleration = v-u/t

avg = 25.9/2.5 = 10.36 m/s²

avg = 10.36 m/s²

Answer 2

Acceleration,
`a=10.4\ m/s^2`

Step-by-step explanation:

Given that,

Initial velocity of the ride, u = 0 (at rest)

Final speed of the ride, v = 22.9 m/s

Time taken, t = 2.2 s

Let a is the acceleration of the passengers on this ride. It can be calculated using the formula of the acceleration as :

`a=(v-u)/(t)`

`a=(22.9-0)/(2.2)`

`a=10.4\ m/s^2`

So, the time required to bring the car to a stop is
`10.4\ m/s^2`. Hence, this is the required solution.