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Motorcycle Kinematic Equation

  • Thread starter 7yler
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A motorcycle officer hidden at an intersection observes a car driven by an oblivious driver who ignores a stop sign and continues through the intersection at constant speed. The police officer takes off in pursuit 1.95 s after the car has passed the stop sign. She accelerates at 4.2 m/s2 until her speed is 108 km/h, and then continues at this speed until she catches the car. At that instant, the car is 1.1 km from the intesection.

I'm not sure how to go about this problem. I don't want the numerical answer, just help with how to approach it.
 

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  • #2
cepheid
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You haven't really stated what the problem is. What are trying to determine?
 
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Sorry, I forgot the rest of the problem. How long did it take for the officer to catch up to the car? How fast was the car traveling?
 
  • #4
cepheid
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Set up equations for the positions of the car and motorcycle vs. time. Let t = 0 be the time at which the motorcycle starts up. That gives you an initial position for the car and its (unknown) constant velocity.

The motorcycle has an initial position and initial velocity of 0. Its motion is divided into two different phases, one with constant acceleration and then another with constant velocity (you have to figure out when this second phase starts up).
 
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I have worked this out and gotten that it takes 27 seconds to catch up with the car, but I'm told that that is wrong. This problem is frustrating me. I really don't understand what I'm doing wrong.
 
  • #6
cepheid
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I have worked this out and gotten that it takes 27 seconds to catch up with the car, but I'm told that that is wrong. This problem is frustrating me. I really don't understand what I'm doing wrong.
Well, we have no way of knowing what you are doing wrong unless if you show us your work. Post your solution here.
 
  • #7
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[tex]v_{0}= 0 km/s[/tex]

[tex]a=0.0042 km/s^{2}[/tex]

[tex]v= 108/3600 km/s[/tex]

[tex]delta x= 1.1 km[/tex]




[tex]108/3600 = 0 + 0.0042(t)[/tex]

[tex]t=108/15.12 for the time of acceleration[/tex]




[tex]delta x=(1/2)(0.0042(108/15.12)^{2}[/tex]

[tex]delta x=0.1071428571 km[/tex]

[tex]1.1 km - 0.1071428571 km = 0.9928571429 km[/tex]

[tex]0.9928571429 km = (108/3600)*t[/tex]

[tex]t=19.85714286 s[/tex]

[tex]19.85714286+(108/15.12)=27s[/tex]
 
Last edited:
  • #8
verty
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You should work in meters and seconds, it is less confusing.
 
  • #9
cepheid
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Sorry for not responding sooner. I've been busy.

[tex]v_{0}= 0~km/s[/tex]

[tex]a=0.0042~km/s^{2}[/tex]

[tex]v= 108/3600~km/s[/tex]

[tex]\Delta x= 1.1~km[/tex]
Good. However, as it has been pointed out, it's probably better if you just consistently use one system of units. In this case, I would convert everything into standard SI units, a.k.a. mks (metre-kilogram-second) units.

By the way, I should mention that you really didn't need the full power of LaTeX for most of these equations. Try using the [ SUB ] [/ SUB ] and [ SUP ] [/ SUP ] tags in order to make regular text subscripted and superscripted respectively (without spaces between the square brackets and the tag text). If you use the buttons marked X2 and X2 above the reply box, it will insert these tags around highlighted text in the reply box automatically. That having been said, I did fix up some of the LaTeX in the quotes of your post. You can click on my versions to see how the source code differs from yours.

[tex]108/3600 = 0 + 0.0042(t)[/tex]

[tex]t=108/15.12~\textrm{for the time of acceleration}[/tex]
Yes, this is the time required for the motorcycle to accelerate up to 108 km/h (30 m/s). However, it is not the total travel time. The rest of the time is spent travelling at a constant speed. You should probably call this time interval t1 or something, to identify it.

[tex]\Delta x=(1/2)(0.0042(108/15.12)^{2}[/tex]

[tex]\Delta x=0.1071428571~km[/tex]

[tex]1.1~km - 0.1071428571~km = 0.9928571429~km[/tex]
This part is also good. You've figured out how far the motorcycle travels while accelerating. The rest of the distance has to be covered at a constant speed.

[tex]0.9928571429~km = (108/3600)t[/tex]

[tex]t=19.85714286~s[/tex]
This calculation step is right, but the answer you got is totally wrong. Try this calculation again. Let's call the result t2

[tex]19.85714286+(108/15.12)=27~s[/tex]
Yes, t1 + t2 is the total travel time of the motorcycle when it is catching up. Hopefully, once you've calculated t2 correctly, you'll get the right answer.

When you work out the speed of the car, don't forget that the car has been travelling 1.95 s longer than the motorcycle.
 
Last edited:

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