A driver driving along a highway at a steady 41 mph (60 ft/sec) sees an accident ahead and slams on the brakes. What constant deceleration is required to stop the car in 200 ft? To find out, carry out the following steps.

constant deceleration required is 9 m/s²

Explanation:

Data provided in the question:

Initial Speed of the driver = 41 mph = 60 ft/s

Stopping distance = 200 ft

Now,

Since the car stops after 200 ft therefore final speed, u = 0 ft/s

from the Newton's equation of motion

we have

v² - u² = 2as

where,

v is the final speed

u is the initial speed

a is the acceleration

s is the distance

thus,

0² - 60² = 2a (200)

or

-3600 = 400a

or

a = - 9 m/s²

here, negative sign means deceleration

Hence,

The constant deceleration required is 9 m/s²