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 :
[tex]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}[/tex]
So, it will cover a distance of 144.64 feet.
