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

Answer:

Vf = 99.7 m/s

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