Acar is moving 14.4 m/s when it hits the brakes. it slows for 78.8 m, which takes 9.92 s. it then accelerates at 1.83 m/s^2 for 4.50 s. what is its final velocity? 2-part motion 1-d kinematics

This is a 2-part motion.
Part 1 of the motion:
Applying equations of motion;
u = 14.4 m/s, s = 78.8 m, t = 9.92 s

s = ut+1/2at^2
78.8 = 14.4*9.92 + 1/2*a*9.92^2
78.8 = 142.848 + 49.2032a
a = (78.8-142.848)/49.2032 = -1.3 m/s^2 (negative sign indicates a deceleration).
Now,
v = u+at
v = 14.4 -1.3*9.92 = 1.49 m/s

Part 2 of the motion:
Final speed in first motion becomes initial velocity in this second motion.
u = 1.49 m/s, a = 1.83 m/s^2, t = 4.50 s

Now,
v = u + at
Then,
v = 1.49 + 1.83*4.50 = 9.72 m/s

The final speed is 9.72 m/s

Answer is c because in exothermic reactions, energy releases.in cellular respiration,atp+h20 (water) = adp+ pi + energyso celullar respiration is exothermic and releases energy