A motorcycle traveling 91.0 km/hr approaches a car traveling in the same direction at 83.0 km/hr. When the motorcycle is 52.0 m behind the car, the rider accelerates and passes the car 16.0 s later. What is the acceleration of the motorcycle (in meters/second^2)?

Answer 1

Easy way: relative motion
relative initial speed: (91-83)km/hr=8km/hr= 2.2222 m/s

distancerelative=virelative*time+1/2 a t^2

52=2.22*16+1/2 a (256)

solve for a.

Harder method:
find the distance the slow car goes in 16 seconds: (velocity*time)
find the distance the fast car must go (above+ 52)
then
that distance=Vi*16 + 1/2 *256 a
check my thinkikng.

Answer 2

.28