A car is traveling along a straight road at a velocity of +37.0 m/s when its eng

Question

A car is traveling along a straight road at a velocity of +37.0 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 +23.9 m/s. The ratio of the average acceleration values is a1/a2 = 1.44. Find the velocity of the car at the end of the initial ten-second interval. *show all work*

A car is traveling along a straight road at a velocity of +37.0 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 +23.9 m/s. The ratio of the average acceleration values is a/a2 1.44. Find the velocity of the car at the end of the initial ten-second interval m/s Additional Materials Section 2.3

Explanation / Answer

Average acceleration is given by:

Avg Acceleration = Change in velocity/time taken

Given that

after 10 sec, avg acceleration is a1,

a1 = (V1 - V0)/(t1 - t0)

V1 = Velovity after 10 sec

V0 = initial Velocity = 37 m/sec

t1 - t0 = 10 sec

So,

a1 = (V1 - 37)/10

Similarly after 5 more second, avg acceleration is a2

a2 = (V2 - V1)/(t2 - t1)

t2 - t1 = 5 sec

V2 = 23.9 m/sec = Velocity after 15 sec

V1 = Velocity after 10 sec, So

a2 = (23.9 - V1)/5

Now Given that

a1/a2 = 1.44

[(V1 - 37)/10]/[(23.9 - V1)/5] = 1.44

[(V1 - 37)/(2*(23.9 - V1))] = 1.44

V1 - 37 = 2.88*(23.9 - V1)

V1*(1 + 2.88) = 2.88*23.9 + 37

V1 = (2.88*23.9 + 37)/3.88

V1 = 27.276 m/sec = 27.3 m/sec

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