Consult Interactive Solution 2.22 before beginning this problem. A car is travel

Question

Consult Interactive Solution 2.22 before beginning this problem. A car is traveling along a straight road at a velocity of +30.0 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 +23.8 m/s. The ratio of the average acceleration values is a1/a2 = 1.79. Find the velocity of the car at the end of the initial ten-second interval.

Explanation / Answer

a2 = 2 then a1 = 1.79 a

for first 10 sec,

vf = v0 - a t (for decelarting motion)

v = 30 - 1.79 a(10)

v = 30 - 17.9 a

then 5 sec,

23.8 = v - a(5)

23.8 = 30 - 17.9a - 5a

a = 0.27 m/^2

so v = 30 - 17.9a

v = 25.2 m/s .........Ans

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