A car is traveling along a straight road at a velocity of +36.0 m/s when its engine cuts out. For the next ten seconds the car slows down, and its average acceleration is a1. For the next five seconds the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the fifteen-second period is +24.5 m/s. The ratio of the average acceleration values is a1/a2 = 1.74. Find the velocity of the car at the end of the initial ten-second interval.

V = 27.07 m/s

Explanation:

Given that a car is traveling along a straight road at a velocity of +36.0 m/s when its engine cuts out. And For the next ten seconds the car slows down, and its average acceleration is a1.

To calculate the acceleration, let use the 1st equation of motion.

V = U - at

As the car is going to rest, V = 0 and a = negative.

V = 36 - a1 × 10

V = 36 - 10a1 (1)

- 10a1 = V - 36

a1 = (36 - V) / 10

Given that For the next five seconds the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the fifteen-second period is +24.5 m/s.

Using the same formula again

V = U - at

24.5 = 36 - 10a1 - a2 × 5

24.5 - 36 = - 10a1 - 5a2

- 11.5 = -( 10a1 + 5a2)

11.5 = 10a1 + 5a2 (2)

But The ratio of the average acceleration values is a1/a2 = 1.74.

a1 = 1.74a2 (3)

Substitute a1 in equation 2

11.5 = 10( 1.74a2) + 5a2

11.5 = 17.4a2 + 5a2

11.5 = 22.4a2

a2 = 11.5/22.4

a2 = 0.5134 m/s^2

Substitutes a2 in equation 3

a1 = 1.74 × 0.5134

a1 = 0.893 m/s^2

Substitutes a1 into equation 1

V = 36 - 10(0.893)

V = 36 - 8.93

V = 27.07 m/s

Therefore, the velocity of the car at the end of the initial ten-second interval is 27.07 m/s

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