Question

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

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

Answer:

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

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 :

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

$a=\dfrac{22.9-0}{2.2}$

$a=10.4\ m/s^2$

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