Question

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.

Answer 1

Answer:

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