Question

A motorcycle 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)?

Answer 1

`a = 0.1137 m/s^2`

Step-by-step 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:

`Xc = d + Vc*t  = Xm = Vm*t + (a*t^2)/(2)`

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

`a = (d+Vc*t-Vm*t)/(t^2)*2` Repacing the values:

`a = 0.1137 m/s^2`