Question

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Answer 1

The car’s velocity at the end of this distance is 18.17 m/s.

Given the following data:

• Initial velocity, U = 22 m/s

• Deceleration, d = 1.4 $m/s^2$

• Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

Final velocity, V = 18.17 m/s

Therefore, the car’s velocity at the end of this distance is 18.17 m/s.

Read more: brainly.com/question/8898885