Question

A car traveling at 45 ft/sec decelerates at a constant 7 feet per second squared. How many feet does the car travel before coming to a complete stop

Answer 1

s = 144.64 feet

Explanation:

Given that,

The initial speed of a car, u = 45 ft/s

Deacceleration of the car, a = -7 m/s²

We need to find the distance covered by the car travel before coming to a complete stop. let it is d.

Final velocity, v = 0

Using the third equation of motion as follows :

`v^2-u^2=2as\\\\s=(v^2-u^2)/(2a)\\\\s=(0^2-45^2)/(2* (-7))\\\\s=144.64\ \text{feet}`

So, it will cover a distance of 144.64 feet.