A car is traveling at a constant velocity of 60.3 miles per hour southward (the

Question

A car is traveling at a constant velocity of 60.3 miles per hour southward (the –x direction) on a straight section of road. At t = 0 and x = 0 it passes an unmarked police car traveling in the same direction at the posted speed limit of 40.0 miles per hour. (a) Using the conversion factors, state both cars’ velocities in SI units (with ± signs to indicate direction). 1.55 seconds after being passed, the police car begins accelerating at 3.50 m/s^2 . (b) At what time and position does the police car pull alongside the first car?

Remember to factor in the 1.55 second gap where the cop car remains at a constant velocity and doesn't accelerate.

Explanation / Answer

Car 1: 60.3 mi/h * 1600 m / 1 mi * 1 h/3600 s = 26.8 m/s

Car Police: 40 mi/h * 1600 m / 1 mi * 1h/3600 s = 17.78 m/s

I have a doubt with part B, but assuming that the police car reach the other one, means that the speed would have to be the same so Vf for this would be 26.8 m/s

a = Vf - Vi / t ---> t = Vf - Vi / t

t = 17.78 - 26.8 / -3.50

t = 2.58 s

but it begins to accelerate after 1.55 s so, the total time would be:

t = 2.58 + 1.55 s = 4.13 s

and the distance would be:

x = Vot + at2 / 2

x = 17.78*4.13 + 3.5*(4.132) / 2

x = 103.28 m

Hope this helps

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