Question

A car is traveling along a straight road at a velocity of +35.0 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 +24.2 m/s. The ratio of the average acceleration values is a1/a2 = 1.78. Find the velocity of the car at the end of the initial ten-second interval.

Answer 1

given initial velocity u = 35 m/s

for ten sec time interval

car slows down with acceleration a1 and let v1 be the velocity

and for next five seconds

v1 is the initial velocity and acceleration is a2

and final speed is v2 = 24.2 m/s

using kinematic equations for first ten seconds time interval

v1 = u+a1*t1

here t1 = 10

v1 = 35+a1*10

v1 = 35+10a1.....(1)

for next five sec time interval

v2 = v1+a2*5

24.2 = v1+5a2...(2)

a1/a2 = 1.78

a2 = a1/1.78 = 0.561 a1

from equations 1 and 2

(1) becomes

v1 = 35+10a1

and 24.2 = v1+5*0.561a1

v1 = 24.2-2.805a1

solving both equations

35+10a1 = 24.2-2.805*a1

a1 = -0.84 m/s^2

from equations

1

v1 = 35-10*0.84

v1 = 35-8.4 = 26.6 m/s