* 22. Mmh A car is traveling along a straight road at a velocity of 36.0 m/s when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is For the next six seconds the car slows down further, and its average acceleration is The velocity of the car at the end of the eighteen-second period is 28.0 m/s. The ratio of the average acceleration values is Find the velocity of the car at the end of the initial twelve-second interval.

Complete question is;

A car is traveling along a straight road at a velocity of 36.0 m/s when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is For the next six seconds the car slows down further, and its average acceleration is The velocity of the car at the end of the eighteen - second period is 28.0 m/s. The ratio of the average acceleration values is a1/a2 = 1.5.

Find the velocity of the car at the end of the initial twelve - second interval.

Your answer

30 m/s

Explanation:

To solve this, we will use Newton's first equation of motion;

v = u + at

Since we are dealing with acceleration ratio, let's make a the subject;

a = (v - u)/t

For the first 12 seconds and initial velocity u = 36 m/s, we have;

a1 = (v - 36)/12

For the remaining 6 seconds, initial velocity is the final velocity from the initial 12 seconds. Thus;

a2 = (28 - v)/6

We are given that a1/a2 = 1.5

Thus;

[(v - 36)/12]/[(28 - v)/6] = 1.5

(6/12)((v - 36)/(28 - v)) = 1.5

(v - 36)/(28 - v) = 1.5 × 12/6

(v - 36)/(28 - v) = 3

v - 36 = 3(28 - v)

v - 36 = 84 - 3v

4v = 120

v = 120/4

v = 30 m/s

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

explanation:

newton's third law, for every action there is an equal and opposite reaction.