Question

A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when the acceleration first began?

Answer 1

88 m/s

Step-by-step explanation:

To solve the problem, we can use the following SUVAT equation:

`v^2-u^2=2ad`

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

`a=-8.0 m/s^2` is the acceleration

Solving for u, we find the initial velocity:

`u=√(v^2-2ad)=√(-2(8.0)(484))=88 m/s`