A motorcycle traveling 95.0 km/hr approaches a car traveling in the same direction at 85.0 km/hr. When the motorcycle is 55.0 m behind the car, the rider accelerates and passes the car 18.0 s later. What is the acceleration of the motorcycle (in meters/second^2)?

Answer 1

Vo = 95000m/3600s = 26.4 m/s.

V = 85000m/3600s = 23.6 m/s.

Vo*t + 0.5a*t^2 = V*t + 55.
26.4*18 + 0.5a*18^2 = 23.6*18 + 55
a = ?