Amotorcycle traveling 95.0 km/hr approaches a car traveling in the same direction at 87.0 km/hr. when the motorcycle is 54.0 m behind the car, the rider accelerates and passes the car 17.0 s later. what is the acceleration of the motorcycle (in meters/second^2)?

Explanation:

Let Vc be the velocity of the car and Vm the velocity of the motorcycle. If we convert their given values, we get:

Vc = 87 km/h * 1000m / 1km * 1h / 3600s = 24.17m/s

Vm = 95 km/h * 1000m / 1km * 1h / 3600s = 26.38m/s

Since their positions are equal after 17s we can stablish that:

Where d is the initial separation distance of 54m. Solving for a, we get:

  Repacing the values:

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

the maxium height is 34m/s