A car is initially at rest at the top of a hill with an incline of 25.0 degrees.

Question

A car is initially at rest at the top of a hill with an incline of 25.0 degrees. The road down the hill is 40.0 m long and ends at the edge of a vertical cliff 100 m in height. 1. What is the acceleration of the car down the hill? 2. How long until the car reached the end of the road? 3. What is the velocity of the car at the end of the road? 4. How long does it take for the car to fall the 100 m to the base of the cliff? 5. How far from the base of the cliff does the car strike the ground? **Hint for question 4: The initial speed in the vertical direction is not equal to zero.

Explanation / Answer

here,

theta = 25 degree

length of incline , l = 40 m

1)

the accelration of the car , a = g * sin(theta)

a = 9.81 * sin(25) = 4.15 m/s^2

2)

let the time taken be t

l = 0 + 0.5 * a * t^2

40 = 0 + 0.5 * 4.15 * t^2

solving for t

t = 4.39 s

3)

the velocity of car , |v| = sqrt(2 * a * l)

|v| = sqrt(2*4.15 * 40) = 18.2 m/s

4)

let the time taken to fall 100 m be t2

h = 100 m = v * sin(theta) * t2 + 0.5 * g * t2^2

100 = 18.2 * sin(25) * t2 + 0.5 * 9.81 * t2^2

solving for t2

t2 = 3.8 s

5)

the distance from the base , x = v * cos(theta) * t2

x = 18.2 * cos(25) * 3.8 m

x = 62.7 m

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