Question

a stunt driver's car is moving at 24.2 m/s along a straight road. The road comes to an abrupt stop 21.2 m before a drop off. The coefficient of friction of the road is 0.712, and the coefficient of friction of the dirt between the road and the drop off is 0.426. What is the minimum distance at which the driver can behin to apply the brakes and still prevent himselff from going over the edge?

Answer 1

Mass of car = m

Initial velocity of car, u = 24.2 m/s

Let velocity of car at the end of road ( start of dirt road ) = w

Final velocity of car = v = 0

Let all vectors in the direction of motion are positive and vice versa

a₁ = acceration of car while on road after applying breaks

a₂ = acceleration of car on dirt path

n₁ = coefficient of friction of road = 0.712

n₂ = 0.426

R = normal reaction acting on car due to road or dirt path = mg  

Let s₁ = distance before the start of dirt path at which driver of car applies breaks

s₂ = distance moved by car before stoping on dirt path = 21.2m

a₁ = -n₁ R/m = - 0.712 mg/m =- 0.712 × 9.81 = - 6.98 ms^-2

a₂ =- n₂ R/m = 0.426 mg/m = -0.426 × 9.81 = - 4.18 ms^-2

On dirt road path ; v² - w² = 2 a₂s₂

w² = v² - 2 a₂s₂ = 0 - 2 × (-4.18)21.2

w = 13.31 ms^-2

on road ; w² - u² = 2 a₁s₁

s₁ = ((13.31)² - (24.2)²)/ 2 × (-6.98) = 29.26 m

Hence driver apolies breaks atleast 29.26 m before the end of road to stop car before the edge.

w² - u² = 2 a₁ s₁

s₁ =( w² - u²)/2 a₁ =