Question

The tires of a car make 82 revolutions as the car reduces its speed uniformly from 89.0 km/h to 59.0 km/h. The tires have a diameter of 0.80 m.

A.) What was the angular acceleration of the tires?

Express your answer using two significant figures.

B.) If the car continues to decelerate at this rate, how much more time is required for it to stop?

Express your answer to two significant figures and include the appropriate units.

C.) If the car continues to decelerate at this rate, how far does it go? Find the total distance.

Express your answer to three significant figures and include the appropriate units.

Answer 1

Radius of the tire, r=(0.8/2)=0.4,

Primary Velocity, u=89 km/h =89*(5/18)= 24.722 m/sec,

Final Velocity, v=59 km/h =59*(5/18) =16.389 m/sec,

Circumference of the tire, L=2*3.14*0.4 =2.512 m,

Therefore, the distance covered by the car after 82 revolution of the tire, s=2.512*82=205.984 m,

A)   From the eqution v²=u²+2fs , we get

Thus the angular acceleration, =f/r =-0.832/0.4=-2.08 rad/sec²

B) From the equation v=u+ft we get

,

As the final velocity is zero when the car stops and we take u=16.389 m/sec as we wanna calculate how much more time is required.

So 19.7 sec more is required to stop the car.

C) From the equation v²=u²+2fs

Thus the total distance covered is 367.514m

16.3892 -24.7222 2 * 205.984 0.832m/sec2