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

Answer:

s = 144.64 feet

Step-by-step 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 :

$v^2-u^2=2as\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{0^2-45^2}{2\times (-7)}\\\\s=144.64\ \text{feet}$

So, it will cover a distance of 144.64 feet.