Question

A roller coaster car starts from rest at the top of a track 35.5 m long and inclined at 40.0° to the horizontal. Assume that friction can be ignored. (a) What is the acceleration (in m/s2) of the car?

Answer 1

Answer:

a = 6.31 m/s²

Explanation:

a) The acceleration of the car can be found using the Newton's second law:

$F = ma$

$W = ma$

$mgsin(\theta) = ma$

Where:

F: is the net force acting on the car

m: is the mass

a: is the acceleration of the car

W: is the weight of the car

θ: is the angle = 40.0°    

g: is the gravity = 9.81 m/s²

Hence, the acceleration is:

$a = gsin(\theta) = 9.81 m/s^{2}*sin(40.0) = 6.31 m/s^{2}$

Therefore, the acceleration of the car is 6.31 m/s².

I hope it helps you!

Answer 2

Complete question:

A roller coaster car starts from rest at the top of a track 35.5 m long and inclined at 40.0° to the horizontal. Assume that friction can be ignored.

(a) What is the acceleration (in m/s2) of the car?

(b) How much time elapses before it reaches the bottom of the track?

Answer:

(a) the acceleration of the car is 6.30 m/s²

(b) elapsed time before it reaches the bottom of the track is 3.36 s

Explanation:

Given;

length of the track, L = 35.5 m

angle of inclination of the track, θ = 40°

initial velocity of the roller coaster, u = 0

Part (a) the acceleration of the car

From Newton's second law;

F = ma = mgsinθ

a = gsinθ

a = 9.8 x sin40

a = 6.30 m/s²

Part (b) elapsed time before it reaches the bottom of the track

Applying kinematic equation;

L = ut + ¹/₂at²

35.5 = 0 + ¹/₂(6.3)t²

35.5 = 3.15t²

t² = 35.5 / 3.15

t² = 11.2698

t = √ 11.2698

t = 3.36 s