Calculating Distance with Motorcycle Patrolman and Car

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To calculate the distance at which a motorcycle patrolman overtakes a car, one must use the equations of motion for both vehicles. The car travels at a constant speed of 124 km/h, while the patrolman accelerates from rest at 6.4 m/s² until reaching a maximum speed of 147 km/h. The time variable for the patrolman is adjusted to account for a 5-second delay in his start. By applying the equations for constant acceleration and distance, the problem can be solved step-by-step. Ultimately, the solution involves rewriting the problem into equations to find the point of overtaking.
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Please help,

A motorcycle patrolman starts from rest at A 5 seconds after a car, speeding at the constant rate of 124 km/h, passes point A. If the patrolman accelerates at the rate of 6.4 m/s2 until he reaches his maximum permissible speed of 147 km/h, which he maintains, calculate the distance s from point A to the point at which he overtakes the car.

Thanks!
 
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hi varsitymsb5! welcome to pf! :wink:

use the standard constant acceleration equations for each vehicle (you'll need t in one to be t+5 in the other) …

show us what you get! :smile:​
 
v=vnot+at
147/3.6=6.4t
t=6.38s
s(car)=124/3.6t'
s(moto)=147/3.6(t'-5)

I'm stuck and not sure what to do next.
 
sorry, I'm not following this at all :confused:

rewrite the words of the question into equations
 
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