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xz_007 [3.2K]
2 years ago
8

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

Physics
1 answer:
Brums [2.3K]2 years ago
8 0

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

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}

<em>Final velocity, V = 18.17 m/s</em>

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

<em></em>

Read more: brainly.com/question/8898885

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Explanation:

Acceleration is equal to the change in velocity over the change in time, or

a=\frac{v_f-v_i}{t} where the change in velocity is final velocity minus initial velocity. Filling in:

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