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/s2 at the moment the motorcycle passes, how much distance will be covered until the car catches up to the motorcycle?
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
xf = xi + vit + at2/2 .
The position of the car is given by the formula
xcar = 22.22t + 5t2/2
and the position of the motorcycle is given by the formula
xmot = 30.56t .
Third, we can equate the last two equations because the car catches up the motorcycle
xcar = xmot
22.22t + 5t2/2 = 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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