A car is moving along the highway at a speed of 80 km/hr. A motorcycle passes the car at a constant speed of 110 km/hr. If the car starts accelerating at 5 m/s^{2} at the moment the motorcycle passes, how much distance will be covered until the car catches up to the motorcycle?

###### Solution.

First, we convert from km/hr to m/s by dividing by 3.6

80 km/hr ≈ 22.22 m/s ; 110 km/hr ≈ 30.56 m/s .

Second, we write the equation for the position as a function of time for the car and for the motorcycle.

The position of the object is given by the formula

x_{f} = x_{i} + v_{i}t + at^{2}/2 .

The position of the car is given by the formula

x_{car} = 22.22t + 5t^{2}/2

and the position of the motorcycle is given by the formula

x_{mot} = 30.56t .

Third, we can equate the last two equations because the car catches up the motorcycle

x_{car} = x_{mot}

22.22t + 5t^{2}/2 = 30.56t

Solving the last equation and we found that

t ≈ 3.34 s .

This is the time required for the car to catch up the motorcycle.

Finally, we substitute the time into the motorcycle’s equation and we get

Distance = 30.56*3.34 ≈ 102.07 m .

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