Question

A blue car moving at constant speed of 10m/s passes a polic car that is at rest. The police car accelerates from rest at 4m/s^2 for 3.0 seconds and then maintains a constant speed. The blue car maintains constant speed for the entire 12 seconds. When does the police car pass the blue car?

Calculations:
Blue Car:
Vi= 10 m/s
Vf = 10 m/s
deltaT = 12 s
deltaD = 120 m
acceleration = 0

Police Car:
Vi = 0m/s
acceleration = 4m/s^2 (for the first 3 seconds)
deltaD (for first three seconds) = 18 m
Vf = 12 m/s

So when the deltaD of the blue car = the deltaD of the police car, the police car is at the same distance as the blue car.

How would I go about doing this?

Answer 1

A blue car moving at constant speed of 10m/s passes a polic car that is at rest. The police car accelerates from rest at 4m/s^2 for 3.0 seconds and then maintains a constant speed. The blue car maintains constant speed for the entire 12 seconds. When does the police car pass the blue car?

The police car accelerates at 4m/s^2 for 3 seconds reaching a final speed of 12m/s^2 and traveling a distance of at^2/2 = 4(3^2)/2 = 18m.

At this 3 second point, the blue car has traveled 3(10) = 30m.

From this point on, the police car will close the existing gap of 12m between the cars at a closing speed of 12 - 10 = 2m/s.

Thus, the police car will overtake, and pass the blue car in 12/2 = 6 seconds or 9 seconds after the blue car started.

Answer 2

ME NO MO