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3 years ago

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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?

1 answer:

Agata [3.3K]3 years ago

6 0

Answer:

88 m/s

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:

$u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s$

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