Question

You are assigned to do some calculations for a movie stunt that involves a car on a straight road. The road, pictured above, has a hill that rises 8.0 m above the flat region. The top of the hill is a circular arc of radius 20 m. You need to determine whether a car traveling under certain conditions will lose contact with the road at the top of the hill. There is a stop sign 50 m from the beginning of the hill. You are to assume that a car of mass 1600 kg accelerates uniformly from rest at the stop sign, has a speed of when it reaches the beginning of the hill, and then coasts with the engine off. Assume energy losses due to friction and air resistance are negligiblea. Calculate the magnitude of the acceleration of the car during the first 50 m
b. Calculate the time it takes the car to reach the beginning of the hill

Answer 1

Answer:

a) a = 2.89 m/s²

b) t ≈ 6 s

Explanation:

Given

h = 8 m

R = 20 m

x = 50 m

v₀ = 0 m/s

v = 17 m/s

m = 1600 kg

a) We can apply the kinematic equation

v² = v₀²+2ax

then

a = (v²-v₀²)/(2x)

⇒ a = ((17 m/s)²-(0 m/s)²)/(2*50 m)

⇒ a = 2.89 m/s²

Now, we can find the normal acceleration as follows

aₙ = v²/R

⇒ aₙ = (17 m/s)²/(20 m)

⇒ aₙ = 14.45 m/s²

Since a < aₙ   (2.89 m/s² < 14.45 m/s²) the car will not reach the top of the hill.

b) We can get the time using the kinematic equation

v = v₀+at

⇒ t = (v-v₀)/a

⇒ t = (17 m/s-0 m/s)/(2.89 m/s²)

⇒ t = 5.88 s ≈ 6 s

Answer 2

Find the solution in the attachment