Question

A car which is traveling at a velocity of 1.6 m/s undergoes an acceleration of 9.2 m/s over a distance of 540 m. How fast is it going

after that acceleration?

Answer 1

Vf = 99.7 m/s

Step-by-step explanation:

In order to find the final velocity of the car, we will use the third equation of motion. The third equation of motion is given as follows:

`2as = V_(f)^(2) - V_(i)^(2)`

where,

a = acceleration of the car = 9.2 m/s²

s = distance traveled by the car = 540 m

Vf = Final Speed of the car = ?

Vi = Initial Speed of the car 1.6 m/s

`2(9.2\ m/s^(2))(540\ m) = (V_(f))^(2)-(1.6\ m/s)^(2)\\V_(f)^(2) = 9936\ m^(2)/s^(2) + 2.56\ m^(2)/s^(2)\\\\V_(f) = \sqrt{9938.56\ m^(2)/s^(2)}`

Vf = 99.7 m/s