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?

Respuesta :

(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

Part a)

As we know that

[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]

now we will have

[tex]v_f = 33.5 m/s[/tex]

[tex]v_i = 0 m/s[/tex]

[tex]\Delta t = 4s[/tex]

now we have

[tex]a = \frac{33.5 - 0}{4} = 8.375 m/s^2[/tex]

Part b)

now for finding the force we will have

[tex]F = ma[/tex]

[tex]F = 45(8.375)[/tex]

[tex]F = 376.875 N[/tex]

so force will be 376.875 N