Three cars are driving at25m/salong the road shown in FIGURE EX8.30. Car B is at the bottom of a hill and car C is at the top. Both hills have a 200m radius of curvature. Suppose each car suddenly brakes hard and starts to skid. What is the tangential acceleration (i.e., the acceleration parallel to the road) of each car? Assume localid="1647757037587" μK=1.0.

Short Answer

Expert verified

The magnitude of tangential acceleration of car A is 9.8m/s2, car B is13m/s2and car C is 6.7m/s2.

Step by step solution

01

Given Infromation

We know that the radiation of curvature of the road is r=200m, speed of the car is v=25m/s, and the coefficient of kinetic friction is μ=1.0.

We have to calculate the acceleration of the car after applying the brake.

02

Explanation

Consider the mass of the car is m. Hence, the weight of the car is w=mg, where g=9.8m/s2is the acceleration due to gravity.

localid="1648323021846" N=mg

The force due to friction is

FA=μN=μmg

Therefore, the acceleration is

aA=μmgm=μg

The acceleration of the car after applying the brake is:

aA=(1.0)9.8m/s2=9.8m/s2.

03

Explanation 

Now car B is at the bottom of the hill. Therefore, the centrifugal force acting on the car will be downwards and the total vertical force acting on the car will be

N-mg=mv2r

N=mg+v2r

Hence, the acceleration is given by

aB=g+v2r

Now substituting the values we have

aB=9.8m/s2+(25m/s)2(200m)

localid="1648323043916" =12.925m/s2

04

Explanation 

Now for the car C is at the top of the hill. Therefore, the centrifugal force will act upward and the total vertical force on the car is:

mg-N=mv2r

Now, similarly the acceleration of the car will be

aC=g-v2r

substituting the values we have

aB=9.8m/s2-(25m/s)2(200m)

=6.7m/s2

05

Final Answer

The magnitude of tangential acceleration of car A is 9.8m/s2, car B is localid="1648323058991" 12.925m/s2and car C is 6.7m/s2.

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