Motorcycle Police Officer Chasing Car

  • Thread starter Ummiya
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  • #1
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This is a practice problem very similar to my homework question just with different values.

1. Homework Statement

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 1.58 s after the car has passed the stop sign. She accelerates at 4.5 m/s2 until her speed is 119 km/h, and then continues at this speed until she catches the car. At that instant, the car is 1.9 km from the intersection.

(a) How long did it take for the officer to catch up to the car?

(b) How fast was the car traveling?

Homework Equations


x(t) = x0 + v0 + 0.5at^2
v(t) = v0 + at

The Attempt at a Solution


So first I would list and convert the values given.
t = 1.58 s
a = 4.5 m/s
vf = 119 km/h ≈ 33 m/s
xf = 1.9 km = 1900 m

I spoke with my professor on this and he said to divide the sections to when the police officer was at rest, when the officer actually takes off, and when the police officer has caught the speeder.

I'm really clueless on how to go through this so I'll just try here.

v(t) = v0 + (4.5)(1.58)
v(t) = 0 + 7.11
v(t) = 7.11 m/s

x(t) = 0 + (7.11)(1.58) + 0.5(4.5)(1.58)^2
x(t) = 16.85 m
 

Answers and Replies

  • #2
SteamKing
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This is a practice problem very similar to my homework question just with different values.

1. Homework Statement

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 1.58 s after the car has passed the stop sign. She accelerates at 4.5 m/s2 until her speed is 119 km/h, and then continues at this speed until she catches the car. At that instant, the car is 1.9 km from the intersection.

(a) How long did it take for the officer to catch up to the car?

(b) How fast was the car traveling?

Homework Equations


x(t) = x0 + v0 + 0.5at^2
v(t) = v0 + at

The Attempt at a Solution


So first I would list and convert the values given.
t = 1.58 s
a = 4.5 m/s
vf = 119 km/h ≈ 33 m/s
xf = 1.9 km = 1900 m

I spoke with my professor on this and he said to divide the sections to when the police officer was at rest, when the officer actually takes off, and when the police officer has caught the speeder.

I'm really clueless on how to go through this so I'll just try here.

v(t) = v0 + (4.5)(1.58)
v(t) = 0 + 7.11
v(t) = 7.11 m/s

x(t) = 0 + (7.11)(1.58) + 0.5(4.5)(1.58)^2
x(t) = 16.85 m

During the 1.58-second interval, the motorcycle is not moving.
 
  • #3
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(Sorry I wrote it as a police car chasing a motorcyclist. It's a motorcycle officer chasing a car.)

Right, so that would make the velocity and everything else zero.
I calculated the point after the officer chases the car which honestly, to me, the numbers don't make sense. I'm sorry I'm really at a loss at this.
 
  • #4
SteamKing
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(Sorry I wrote it as a police car chasing a motorcyclist. It's a motorcycle officer chasing a car.)

Right, so that would make the velocity and everything else zero.
I calculated the point after the officer chases the car which honestly, to me, the numbers don't make sense. I'm sorry I'm really at a loss at this.
No need to panic here. Take the calculation in steps.

The police officer accelerates at 4.5 m/s2 from rest to a speed of 119 kph (or 33 m/s).

How long does it take to do this?

How far does the motorcycle travel during this time?

Start keeping track of the time elapsed and the distance traveled.
 
  • #5
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So using v(t) = v0 + at
33 = 0 + 4.5t
t = 7.3 s
It took the officer 7.3 seconds to reach the speed of 119 k/h (33 m/s).

And x(t) = x0 + v0t + 0.5at^2
x(t) = 0 + 33(7.3) + 0.5(4.5)(7.3)^2
x(t) ≈ 360
The officer has traveled 360 meters during this time.
 
  • #6
SteamKing
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So using v(t) = v0 + at
33 = 0 + 4.5t
t = 7.3 s
It took the officer 7.3 seconds to reach the speed of 119 k/h (33 m/s).

And x(t) = x0 + v0t + 0.5at^2
x(t) = 0 + 33(7.3) + 0.5(4.5)(7.3)^2
x(t) ≈ 360
The officer has traveled 360 meters during this time.

The calculation looks good so far.

After reaching 119 kph, the police officer travels at this speed until she is 1900 meters from her starting point. How long does it take to reach this distance, given that the officer has already traveled 360 meters while accelerating?
 
  • #7
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Okay so then using the same equation again with zero acceleration now because she is maintaining the speed.
1900 = 360 + 33t
1540 = 33t
t = 46.7 s
It took the officer 46.7 seconds to reach the 1900 meter position from the 360 meters position.
 
  • #8
SteamKing
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Okay so then using the same equation again with zero acceleration now because she is maintaining the speed.
1900 = 360 + 33t
1540 = 33t
t = 46.7 s
It took the officer 46.7 seconds to reach the 1900 meter position from the 360 meters position.

What is the total elapsed time for the police woman to catch the car?

If the car is traveling at a constant speed during the chase, what is the speed of the car?
 
  • #9
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t(total) = 1.58 + 7.3 + 46.7 = 55.58 seconds

33 m/s?
 
  • #10
SteamKing
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t(total) = 1.58 + 7.3 + 46.7 = 55.58 seconds

33 m/s?
Did you calculate 33 m/s, or did you just assume the car was traveling 119 kph, like the motorcycle?
 
  • #11
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I assumed that when you said that the speed of the car was constant throughout the whole chase and that the police woman had to catch up to that speed.

Also, the answer for how long it took for the police woman to catch up to the car says 62.7 seconds in my practice. I don't know which calculations may have gone wrong that I got 55.58 seconds.
 
  • #12
SteamKing
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I assumed that when you said that the speed of the car was constant throughout the whole chase and that the police woman had to catch up to that speed.

I think the problem meant for you to calculate the speed of the car, given what the police woman had to do to catch up with it.

Also, the answer for how long it took for the police woman to catch up to the car says 62.7 seconds in my practice. I don't know which calculations may have gone wrong that I got 55.58 seconds.

It's not clear to me how the practice calculates a time of 62.7 seconds.
 
  • #13
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Then should I be setting two equations equal to each other?

There's this note, it's online practice so after you get the answer wrong you get this: "The time the speeder continues at constant velocity is sum of the times required for the three constant-acceleration segments of the police officer's chase. The final time and position of the police officer for each segment are equal to the initial time and position of the police officer for the next segment."

Which I think we already understood as breaking it up into segments.
 
  • #14
SteamKing
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Then should I be setting two equations equal to each other?

What two equations?
 
  • #15
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I'm not sure, but I assume from this: "The time the speeder continues at constant velocity is sum of the times required for the three constant-acceleration segments of the police officer's chase."
 
  • #16
SteamKing
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I'm not sure, but I assume from this: "The time the speeder continues at constant velocity is sum of the times required for the three constant-acceleration segments of the police officer's chase."
That's the calculation you have already performed, namely:
1. Reaction time of officer: 1.58 sec.
2. Acceleration of motorcycle to 119 kph: 7.3 sec.
3. Motorcycle travels at constant speed until car is caught: 46.7 sec.
 
  • #17
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Right. So what can I do at this point? The answer would still be wrong (55.58 seconds). I tried again by not rounding the numbers and keep them true to their value but still didn't get the answer in the practice.
 
  • #18
you should write an equation to describe the vehicle travelling at constant velocity. That should be easy if you know the speed of the vehicle.
you also have to make an equation for the motorcycle police officer which would include the initial rest period, the acceleration period and the constant velocity section (which would have to be greater than the constant velocity of the vehicle). The key piece of information is that the motorcycle DOES catch the car so that means there is a TIME in seconds that is the same number for the travelling car and the pursuing motorcycle. YES it is a simultaneous equation problem or could also be explained as a Speed versus Time graph which shows the motorcycle and car to have equal displacement when the motorcycle catches the speeding car.
 
  • #19
haruspex
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So using v(t) = v0 + at
33 = 0 + 4.5t
t = 7.3 s
It took the officer 7.3 seconds to reach the speed of 119 k/h (33 m/s).

And x(t) = x0 + v0t + 0.5at^2
x(t) = 0 + 33(7.3) + 0.5(4.5)(7.3)^2
x(t) ≈ 360
The officer has traveled 360 meters during this time.
How did v0 change from 0 in the calculation to find t to being 33 in the calculation to find x?
 

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