Question

A Roller coaster will accelerate its riders from rest to 33.5m/s in 4.00 seconds.

(a) What is the magnitude of the average acceleration of a rider?


(b) What is the average net force on a 45.0 kg rider during these 4.00 seconds?

Answer 1

Part a)

As we know that

`a = (v_f - v_i)/(\Delta t)`

now we will have

`v_f = 33.5 m/s`

`v_i = 0 m/s`

`\Delta t = 4s`

now we have

`a = (33.5 - 0)/(4) = 8.375 m/s^2`

Part b)

now for finding the force we will have

`F = ma`

`F = 45(8.375)`

`F = 376.875 N`

so force will be 376.875 N

Answer 2

(a)

v₀ = initial velocity of the roller coaster = 0 m/s

v = final velocity of the roller coaster = 33.5 m/s

t = time of travel = 4 seconds

a = magnitude of average acceleration

magnitude of average acceleration is given as

a = (v - v₀)/t

a = (33.5 - 0)/4

a = 8.375 m/s²

b)

m = mass of the rider = 45 kg

a = magnitude of acceleration of the rider = 8.375 m/s²

F = average net force on the rider

average net force on the rider is given as

F = ma

F = 45 x 8.375

F = 376.875 N