This physics problems is crazy difficult?

Question

A car is traveling along a straight road at a velocity of +31.0 m/s when its engine cuts out. For the next ten seconds, the car slows down, and its average acceleration is a bar1. For the next five seconds, the car slows down further, and its average acceleration is a bar2. The velocity of the car at the end of the fifteen-second period is +22.3 m/s. The ratio of the average acceleration values is a bar1/a bar2 = 1.20. Find the velocity of the car at the end of the initial ten-second interval.

Answer 1

From instant t = 0 to t = 10, the velocity change is:

(Δv)1 = a1.(10 - 0) = 10.a1

From instance t = 10 to t = 15, the velocity further change is:

(Δv)2 = a2.(15 - 10) = 5.a2

(Δv)1 + (Δv)2 = 22.3 - 31.0 = - 8.7 m/s

also:

(Δv)1 + (Δv)2 = 10 a1 + 5 a2

So:

10 a1 + 5 a2 = - 8.7 (1)

On the other hand:

a1 / a2 = 1.20

a1 - 1.2a2 = 0 (2)

Solve the system of equations (1) and (2) for a1 and a2:

10 a1 + 05 a2 = - 8.7

10 a1 -- 12 a2 = 0

a2 = -8.7/17 m/s^2

a1 = -8.7*1.2/17 m/s^2

(Δv)1 = 10 a1 = -10*8.7*1.2/17 = - 6.1 m/s

Velocity after 10 initial seconds:

v(10) = v(0) + (Δv)1 = 31.0 -- 6.1 = 24.9 m/s