Kinematics: Basic Homework Question on Cop and Car

[Femme_physics]

Homework Statement

A traffic officer riding on a bike starts his motion at point A, 3 seconds after the a private car went by him. The private car is traveling in constant velocity of 120 km/h.

The officer is accelerating at 5 meters per second squared, till a speed of 150 km/h, which is his allowed maximum velocity, and continues driving in that speed.

A) Calculate the time it takes the officer to catch up with the car.
B) Calculate the distance from point A till the point where the police officer moves ahead of the car.http://img810.imageshack.us/img810/7407/600d.jpg

Homework Equations

Kinematics

The Attempt at a Solution

I'll start with a disclaimer, I was a bit confused about how to go about solving that with my lack of experience solving kinematics problems. In fact, I'll continue with a confession -- I glimpsed at the solution manual and I saw that the solution starts by calculating when the cop reaches max velocity, and the distance he traveled when he reaches max velocity.
Now, how am I supposed to know if the cop didn't already pass the car BEFORE reaching max velocity? It may be intuitive, but what if velocity was 50, and not 5?

This may be a long thread since, though I understand the kinematics equations and how to manipulate them, I am highly inexperienced in picking the right ones to get what I need. So, the question above is my first question.

 

[I like Serena]

This may be a long thread.

Dooly noted! :)

Homework Statement

A traffic officer riding on a bike starts his motion at point A, 3 seconds after the a private car went by him. The private car is traveling in constant velocity of 120 km/h.

The officer is accelerating at 5 meters per second squared, till a speed of 150 km/h, which is his allowed maximum velocity, and continues driving in that speed.

A) Calculate the time it takes the officer to catch up with the car.
B) Calculate the distance from point A till the point where the police officer moves ahead of the car.

The Attempt at a Solution

I'll start with a disclaimer, I was a bit confused about how to go about solving that with my lack of experience solving kinematics problems. In fact, I'll continue with a confession -- I glimpsed at the solution manual and I saw that the solution starts by calculating when the cop reaches max velocity, and the distance he traveled when he reaches max velocity.
Now, how am I supposed to know if the cop didn't already pass the car BEFORE reaching max velocity? It may be intuitive, but what if velocity was 50, and not 5?

Yes, you don't know yet if the cop passed the car before reaching max velocity.
Let's hope that when we work it out, we can see something that might tell us.
So we'll keep an eye out for stuff like that! :)
(That's always a good idea! )

The first stage of the problem, however, is figuring out the initial situation when the cop starts accelerating.
How far is the car ahead at that time?

 

[Femme_physics]

Yes, you don't know yet if the cop passed the car before reaching max velocity.
Let's hope that when we work it out, we can see something that might tell us.
So we'll keep an eye out for stuff like that! :)

Good!

The first stage of the problem, however, is figuring out the initial situation when the cop starts accelerating.
How far is the car ahead at that time?

The car is, well, let's see, it drives in 120 km/h, so in meters per second it's 120/3.6 = 33.333333 [m/s]

Now we multiply his speed by 3 seconds

33.333333 x 3 = 100 [m/s]

It's 100 meters ahead.

That's an interesting fact, actually.

Hmm...let me see something here.
If I calculate how far the officer gets just when he quits accelerating, I can use that same time to answer my original question.

Okay, so:

150/3.6 = 0 + 5t
t = 8.333333

So it takes the biker 8.33333 of acceleration to get to max speed.

Which means he's gotten to a distance of

x = 0 + 0 x 8.333 + 5x8.3333²/2

x = 173.609

That's the distance.

Now let's plug that to the car which started 3 seconds later

173.609 = 0 + 120 x 11.3333

[I added 3 seconds since the car ]

173.609 does not equal 1359.996!

The car was way ahead then :)

Now, how to do the comparison between the two to see where the officer sped ahead of the car, that may take a little more brain churning...

 

[I like Serena]

Whooaah!
This is so flawless, there will be very little left to this thread!

[edit] Yep, the car was way ahead! [/edit]

So you're entering the third stage of the problem, where both the cop and the car drive at a constant speed, and they start out with a known difference in distance.

The easiest way to do this is to look at the difference in speeds and calculate the time it takes to bridge the difference in distance.
And then use this time to calculate the extra distance the cop has covered.

 

[Femme_physics]

Whooaah!
This is so flawless, there will be very little left to this thread!

lol & thanks! :) Glad to get it right!

So you're entering the third stage of the problem, where both the cop and the car drive at a constant speed, and they start out with a known difference in distance.

So I have distance 11.3333 seconds after the car first blew pass the officer. I know they both have constant velocity, which means the "a" term drops. I also know their initial velocity. zero and 120 just from the facts. So, the only thing I need to relate is time. I'll plug in all that information to a single equation to see how to relate them..as soon as I have the eq I'll post back here :)

 

[I like Serena]

lol & thanks! :) Glad to get it right!

I'm starting to get the impression that you like dynamics!

So I have distance 11.3333 seconds after the car first blew pass the officer. I know they both have constant velocity, which means the "a" term drops. I also know their initial velocity. zero and 120 just from the facts. So, the only thing I need to relate is time. I'll plug in all that information to a single equation to see how to relate them..as soon as I have the eq I'll post back here :)

Huh?
A distance in seconds?
An initial speed of zero? At the start of the third phase that should be 150 km/h.

Well, I'll let you puzzle a bit on it, and I'll see you back here. :)

 

[Femme_physics]

Huh?
A distance in seconds?

No no, I mean I have the distance after that certain amount of seconds passed.

An initial speed of zero? At the start of the third phase that should be 150 km/h.

Oh, that's right. Good you caught it before I plugged it into the equation.

I'll puzzle it for a bit

 

[Femme_physics]

My idea seemed so perfect, I'm not sure why I got the wrong answer!

t1 = 8.33333
t2 = 11.3333
xo1 = 173.609
xo2 = 1360
vo1 = 150
vo2 = 120
v1 = 150
v2 = 120


Xo₁ + V₁t = Xo₂ +V₂t
30t = 1186.391
t = 39.546


The answer in the manual is 35.8 sec

 

[I like Serena]

My idea seemed so perfect, I'm not sure why I got the wrong answer!

t1 = 8.33333
t2 = 11.3333
xo1 = 173.609
xo2 = 1360
vo1 = 150
vo2 = 120
v1 = 150
v2 = 120Xo₁ + V₁t = Xo₂ +V₂t
30t = 1186.391
t = 39.546The answer in the manual is 35.8 sec

I backtracked your calculation and I'm afraid your previous calculation was not flawless after all. :(

Your distance for the car of 1360 meters is not right.
You forgot to plug in the conversion factor for km/h...

[edit]You forgot the conversion factor again in your last calculation.

Tip: convert everything to SI units (meters, seconds, kilograms) before even starting to solve a problem.
That way you won't make mistakes with conversion factors.[/edit]

 

[Femme_physics]

I'm starting to get the impression that you like dynamics!

Yea, it's so much more...how should I put it... dynamic, than statics :)

(really I should be slapped for that)

I backtracked your calculation and I'm afraid your previous calculation was not flawless after all. :(

Your distance for the car of 1360 meters is not right.
You forgot to plug in the conversion factor for km/h...

[edit]You forgot the conversion factor again in your last calculation.

Tip: convert everything to SI units (meters, seconds, kilograms) before even starting to solve a problem.
That way you won't make mistakes with conversion factors.[/edit]

Good idea.

So, it should be instead of

173.609 = 0 + 120 x 11.3333

173.609 = 0 + (120/3.6) x 11.3333

173.609 < 377.777

Good, so the car is still ahead of the bike. Except this time the distance is much smaller than it first appeared.

So let's rewrite this


t1 = 8.33333
t2 = 11.3333
xo1 = 173.609
xo2 = 377.777
vo1 = 41.6667
vo2 = 33.3333
v1 = 41.6667
v2 = 33.3333



Xo₁ + V₁t = Xo₂ +V₂t

173.609 + 41.6667t = 377.777 +33.3333t

t = 24.5

Hmm... Okay, I've tried backtracking but I can't tell what it is. I got the SI units, I used the same equation as before, I plugged respectively the distances and velocities...

What can it be?

 

[I like Serena]

Yea, it's so much more...how should I put it... dynamic, than statics :)

(really I should be slapped for that)

*slap*
How did that feel for you?

Anyway, it's about motorbikes... and I *like* motorbikes!

Good idea.

So, it should be instead of

173.609 = 0 + 120 x 11.3333

173.609 = 0 + (120/3.6) x 11.3333

I let it slide before, but actually this should be:

377.778 = 0 + (120/3.6) x 11.3333

173.609 < 377.777

Good, so the car is still ahead of the bike. Except this time the distance is much smaller than it first appeared.

Yep!

t = 24.5

Hmm... Okay, I've tried backtracking but I can't tell what it is. I got the SI units, I used the same equation as before, I plugged respectively the distances and velocities...

What can it be?

Well, this is the time starting when the cop was at full speed.
How much time had already passed at that time?

 

[Femme_physics]

*slap*
How did that feel for you?

*rubs cheek* mildly enjoyable, actually...

I let it slide before, but actually this should be:

377.778 = 0 + (120/3.6) x 11.3333

Well yes, that's why I switched sign from equal to "bigger than". I was just making a comparison

Well, this is the time starting when the cop was at full speed.
How much time had already passed at that time?

I see...

So, first I do

X = 173.609 + 41.6667 x 3
X = 298.61

Xo₁ + V₁t = Xo₂ +V₂t

298.61 + 41.6667t = 377.777 +33.3333t
t = 9.5

Hmm... now I'm even more far off than before

 

[I like Serena]

I see...

So, first I do

X = 173.609 + 41.6667 x 3
X = 298.61

Xo₁ + V₁t = Xo₂ +V₂t

298.61 + 41.6667t = 377.777 +33.3333t
t = 9.5

Hmm... now I'm even more far off than before

I've lost you here. What did you do here?

Let me make an inventory.

First the cop waits 3 seconds.
Then he accelerates for 8.333 seconds.
Then he overtakes the car in 24.5 seconds.

So how much time has passed?

 

[Femme_physics]

I've lost you here. What did you do here?

Sorry for the lack of annotation, I wrote a longer post but it got deleted so I got cranky and wrote it shorter :P

What I did is...oh...nevermind. I see my mistake. All I'm trying to do is match their times.

Then he accelerates for 8.333 seconds.
Then he overtakes the car in 24.5 seconds.

So how much time has passed?

OHHHHH! So I was right originally, I see now... *smacks forehead* .. Thanks for that bone ;) Damn it!

Must...learn...more... ;D

I'll get to question B soon.

 

[Femme_physics]

8.33333 + 24.5 = 32.83

Hmm...still not the answer

 

[I like Serena]

8.33333 + 24.5 = 32.83

Hmm...still not the answer

And what if you add 3 seconds?

 

[Femme_physics]

Wait a second, I thought the 3 seconds are taken into consideration in this equation ->

377.778 = 0 + (120/3.6) x 11.3333Instead of seeing how far the car has gotten at 8.333 seconds, we see how far it's gotten in 11.33333 seconds. Therefor, we eliminate the need to use the 3 seconds thingy. Or so I thought. Hmm... shouldn't we have done then

X = 0 + (120/3.6) x 8.3333

And add the 3 seconds later like we did now? I'm a bit confused.

 

[Femme_physics]

Wait... I'll think about it.. I'll do what my classmate calls "reverse engineering" :)

 

[I like Serena]

Wait... I'll think about it.. I'll do what my classmate calls "reverse engineering" :)

That's good practice, so I'll let you to it, for now...

 

[Femme_physics]

I'm looking at this equation

Xo₁ + V₁t = Xo₂ +V₂t

173.609 + 41.6667t = 377.777 +33.3333t

t = 24.5

This equation and result is for the time AFTER 11.3333 seconds.

I see now and understand perfectly :)

I'll get to B later. Much appreciated ILS!

 

[I like Serena]

I'm looking at this equation

Xo₁ + V₁t = Xo₂ +V₂t

173.609 + 41.6667t = 377.777 +33.3333t

t = 24.5

This equation and result is for the time AFTER 11.3333 seconds.

I see now and understand perfectly :)

I'll get to B later. Much appreciated ILS!

 

[Femme_physics]

Got question B :)

X = 0 + 120/3.6 x 35.83

X = 1194.33333

 

[I like Serena]

Got question B :)

X = 0 + 120/3.6 x 35.83

X = 1194.33333

Yep!

And that was also the easiest way to calculate it!
It's fun, n'est-ce pas?

 

[Femme_physics]

Very. Simple and logical, too! :)

Well, you make it seem like it, anyway!

 

[Femme_physics]

Can this problem be solved with calculus?

 

[noleguy33]

You can 'dress' it up with calculus, but with constant acceleration, it is much easier to use algebra.

 

[Femme_physics]

I see

 

[I like Serena]

Hey Fp!
Mooooooooorning!

It's as noleguy33 says (I read "noteguy" at first).

However, I would like to share a calculus statement with you.

You already know that
$$x = x_i + v_i t + \frac 1 2 a t^2$$

A funny thing to note is that if you take the derivative, you'll get:
$$x' = v_i + a t$$

Does that look familiar?

 

[Femme_physics]

Hmm...well it doesn't appear anywhere here

http://www.scribd.com/doc/57839921/All-Dynamics-Formulae

It's just slightly similar to

Vf = Vi + at

 

[I like Serena]

Yes. That is the one I meant.

So the derivative of x to t equals the velocity v!

Or in other words, the change in distance x per unit of time, equals the velocity.

 

[Femme_physics]

Oh yea, I remember that from a certain calculus teacher.

The derivative of distance over time is velocity, the derivative of velocity over time is acceleration, and the derivative of acceleration over time is jerk!

 

[flyingpig]

I just did the problem myself and it took me about 20 mins...and that's only part a

This is NOT a basic kinematics problem...

That or my brain is getting rusty from pure math lol.

 

[Femme_physics]

Well it's pretty linear (without calculus - did u try it with calculus?), a little long but not too long and/or difficult IMO.

 

[I like Serena]

Oh yea, I remember that from a certain calculus teacher.

The derivative of distance over time is velocity, the derivative of velocity over time is acceleration, and the derivative of acceleration over time is jerk!

A certain calculus teacher?
I remember "jerk" from XKCD!

 

[flyingpig]

Honestly it's even more difficult to use calculus to this problem