Question

The standard test to determine the maximum lateral acceleration of a car is to drive it around a 200-ft-diameter circle painted on a level asphalt surface. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. A 3000 lb car is traveling at 25 mi/hr when the driver applies the brakes, and the car continues to move along the circular path. What is the maximum deceleration possible if the tires are limited to a total horizontal friction force of 2400 lb?

Answer 1

Answer:

$a=25.736\ lb.ft.s^{-2}$ is the maximum deceleration from this top speed keeping-up with the grip of friction.

Explanation:

Given:

diameter of the track, $d=200\ ft$

mass of the car, $m=3000\ lb$

speed of the car, $v=25\ mi.hr^{-1}=36.6667\ ft.s^{-1}$

maximum horizontal frictional force between the surfaces, $f=2400\times 32.17=77208\ lb.ft.s^{-2}$

Now the maximum speed attained by the car according to the frictional force:

$f=m.\frac{v^2}{r}$ also $f=m.a$

where:

• a = acceleration; $\frac{v^2}{r} =a$

• $r=\frac{d}{2} =100\ ft$

$77208=3000\times \frac{v^2}{r}$

$a=25.736\ lb.ft.s^{-2}$ is the maximum deceleration from this top speed keeping-up with the grip of friction.