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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?

User Kchomski
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2 Answers

4 votes

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

User Pseudoanime
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8.3k points
3 votes

(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

User Gadu
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8.5k points