Solve Kinematics Confusion: Motorist vs Police Car

[Exuro89]

Homework Statement

A motorist traveling at a constant speed of 160 km/h in a 50-km/h speed zone passes a parked police car. Three seconds after the car passes, the police car starts off in pursuit. The policeman accelerates at 2m/s^2 up to a speed of 30 m/s, and then continues at this speed until he overtakes the speeding motorist. How long from the time he started does it take the police car to overtake the motorist? The motorist continues at a constant speed during this process.

Homework Equations

X=X_o+V_o*t+.5at^2

The Attempt at a Solution

I'm somewhat confused with how the cop car is suppose to pass the motorist. It says that the motorist is going at a constant 160km/h which is equal to 44.44m/s correct? It then states that when the police car is passed the police car beings to accelerate at 2m/s^2 until a velocity of 30m/s. How could the police car even catch up to the motorist if it isn't going faster than the motorist? Am I reading this problem incorrectly?

 

[SammyS]

Yes, 160 km/hr = 44.44 m/s.

Are you sure it's not 60 km/hr rather than 160 km/hr?

 

[Exuro89]

Yes I'm sure. It was multiple choice and apparently the answer is a negative...

 

[SammyS]

If you blindly plug into the right equations, of course the answer would be negative. Algebra has no way to know how nonsensical that would be.

As you noticed, there is no time in the future when the car would overtake the motorist.

BTW: I suspect that the negative answer is wrong too. It assumes the policeman is traveling backwards during the time the problem states he is stopped.

 

[rwisz]

I would approach this problem by first clarifying the initial conditions and assumptions. It is stated that the motorist is traveling at a constant speed of 160 km/h and the police car is initially parked. From this, we can assume that the police car is stationary and the motorist is approaching it at 160 km/h.

Now, the police car starts off in pursuit three seconds after the motorist passes. This means that the police car starts moving three seconds later than the motorist and has to catch up to the motorist's position at that time.

The police car accelerates at 2m/s^2 until it reaches a speed of 30 m/s. From this, we can calculate the time it takes for the police car to reach this speed using the equation v = u + at, where u is the initial velocity (0 m/s) and v is the final velocity (30 m/s). This gives us a time of 15 seconds.

Once the police car reaches a speed of 30 m/s, it continues at this speed until it overtakes the motorist. We can calculate the distance traveled by the police car during this time using the equation s = ut + 0.5at^2, where u is the initial velocity (30 m/s), t is the time taken (15 seconds), and a is the acceleration (0 m/s^2, since the police car is now moving at a constant speed). This gives us a distance of 450 meters traveled by the police car.

Now, since the police car has to catch up to the motorist's position at the time it started moving (three seconds after the motorist passed), we can calculate the distance traveled by the motorist in those three seconds using the equation s = ut + 0.5at^2, where u is the initial velocity (160 km/h or 44.44 m/s), t is the time taken (3 seconds), and a is the acceleration (0 m/s^2, since the motorist is traveling at a constant speed). This gives us a distance of 133.33 meters traveled by the motorist.

Therefore, the total distance the police car needs to travel to overtake the motorist is 450 meters + 133.33 meters = 583.33 meters.

To calculate the time it takes for the police car to overtake the motorist, we