Police car problem. i need it in an hour

[iffanytay]

An unmarked police car, traveling a constant 87.0km/hr, is passed by a speeder traveling 134km/hr. Precisely 1.11s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 1.94m/s2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at constant speed)?

anything!

 

[Andrew Mason]

An unmarked police car, traveling a constant 87.0km/hr, is passed by a speeder traveling 134km/hr. Precisely 1.11s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 1.94m/s2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at constant speed)?

anything!

Do a distance - time graph for each of the vehicles. What is the relationship between the co-ordinates (distance and time) of the two cars when the police car overtakes the speeder?

AM

 

[iffanytay]

thank you ill try that. does anyone know how to do it using the x=xo+vot+1/2at^2 equation?

 

[LowlyPion]

An unmarked police car, traveling a constant 87.0km/hr, is passed by a speeder traveling 134km/hr. Precisely 1.11s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 1.94m/s2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at constant speed)?

anything!

Rather than looking for an equation, why don't you start by figuring all the things you know?

First of all convert everything to m/sec before you get tripped up by that.
87 km/hr = ? m/s
134 km/hr = ? m/s

Next how far past the police does the speeder get before the police car starts up? You need that distance

Starting from t=0 when the police steps on the gas, what is the distance the police will go considering his initial velocity and constant acceleration 1.94? This will be an equation that describes his X-position as a function of time.

The speeder has his own equation that describes his position as a function of time from the same t=0. What is it? (Remember he has the 1.11 sec * his speed head start)

You should have two equations that describe each car's position as a function of time. When they are equal to each other - guess what that = caught.

 

[iffanytay]

I already converted and I've set the individual problems up for each car so many times but somehow i just get lost. I don't know what to do once i get to that point.

 

[vociferous]

thank you ill try that. does anyone know how to do it using the x=xo+vot+1/2at^2 equation?

Is this a calculus based course or a non-calculus based course?

 

[iffanytay]

its non-calculus.

 

[iffanytay]

can someone just walk me through it and explain the whole thing. that would be the most helpful.

 

[LowlyPion]

I already converted and I've set the individual problems up for each car so many times but somehow i just get lost. I don't know what to do once i get to that point.

Write out your equation here and we can help you.

 

[vociferous]

Since this is a non-Calculus based course, you have already been given the equations you need to solve it, as you already listed: x=xo+vot+1/2at^2

This is a standard position equation. What you need to do is take the data given to you in the problem, create a system of equations using that position equation; one for the police car, one for the speeder, then see where they intersect. Then you will have x and t (position and time). You will have to modify the equation for the policemen to adjust for the 1.11 second delay.

 

[iffanytay]

if anyone has the answer please can i have it. i keep getting so lost when i try to set the equations equal to each other and it is due really soon.

 

[LowlyPion]

if anyone has the answer please can i have it. i keep getting so lost when i try to set the equations equal to each other and it is due really soon.

Sorry, can't give you the answer. I would help you solve it, if you had ever posted your work. Besides, I already showed you how to go about it.

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