Calculating Time and Velocity in One-Dimensional Kinematics

[ScienceGeek]

I am reviewing General Physics for the MCAT and have a question regarding kinematics in one dimension. The question reads as follows:

A speeding motorist traveling 120 km/h passes a stationary police officer. The officer immediately begins pursuit at a constant acceleration of 10.0 km/h/s. How much time will it take for the police officer to reach the speeder, assuming that the speeder maintains a constant speed? How fast will the police officer be traveling at this time?
I am unsure how to solve this problem, conceptually speaking. I converted the mixed units to units compatible with m/s. My calculated values are as follows:

Vmotorist = 33 m/s
V0cop = 0 m/s
V cop = ?
Acop = 2.8 m/s2
T = ?

I have used the following kinematics equations:
V2 = V02 + 2a∆X
V = V0 + aT
X = X0 + V0T + ½ aT2

I calculated the time it would take the cop to reach 33 m/s and the distance at that time:
V= V0 + aT X = X0 + V0T + ½ aT2
33 m/s = (2.8 m/s2)*t = ½ * 2.8 * 11.82
T = 11.8 s = 194.9 m

I also calculated the difference in distance between the motorist and the cop at this point (194.5 m) however I am unsure as to whether or not these values are even relevant. Any assistance with the direction I should go into solve these questions would be greatly appreciated. Thank you!

 

[Doc Al]

Try this. Write separate formulas for the position of each as a function of time. When will those positions be equal?

 

[cristo]

I am reviewing General Physics for the MCAT and have a question regarding kinematics in one dimension. The question reads as follows:

A speeding motorist traveling 120 km/h passes a stationary police officer. The officer immediately begins pursuit at a constant acceleration of 10.0 km/h/s. How much time will it take for the police officer to reach the speeder, assuming that the speeder maintains a constant speed? How fast will the police officer be traveling at this time?
I am unsure how to solve this problem, conceptually speaking. I converted the mixed units to units compatible with m/s. My calculated values are as follows:

Vmotorist = 33 m/s
V0cop = 0 m/s
V cop = ?
Acop = 2.8 m/s2
T = ?

Ok, good.

Now, you want to find the time at which the cop will catch the speeder. At this time the displacement from the starting point of the cop of each car will be equal (otherwise, he wouldn't have caught him!) You tried to say that when the cop catches the speeder, both of their velocities are the same-- this is incorrect.

So, can you write an expression for the distance covered by each car? [Hint: leave x and t as v unknown variables]

If you can do this, then you will have set up two equations in two unknowns; you can solve these for x and t.

 

[cristo]

cuddlesome: here at PF we try to give tutorial advice, and guide students through questions. Please refrain from posting full solutions in future, as per the PF guidelines to homework help.

 

[ScienceGeek]

Thank you all! I set the distance equations equal to each other, eliminating X as a variable essentially and solving for t. I checked my answers with those provided in the back of the book and they were correct.