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 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Mar 20th 2011, 01:42 AM #1 Member   Join Date: Oct 2009 Posts: 58 Kinematics Two cars face each other on a horizontal road. Car A starts from rest at t=0 amd travels with a constant acceleration of 60m/s^2, until it reaches a speed of 80ft/s. Afterwards it maintain the same speed.Also after t=50 sec , Car B located 6000 ft down the road is traveling towards A with a constant speed of 60 ft/s. Determine the distance traveled by A when they pass each other. We've done this problem , having both cars launched at the same time. What are the changes that might occur. Thanks in advance.
 Mar 20th 2011, 09:56 AM #2 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 If t is the time, you'll use t for car A and for car B, you'll use (t-50) for its time, so that when t = 50, car B has got time 0, meaning that it's only now starting to move from a said point. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
Mar 20th 2011, 11:21 AM   #3

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 Originally Posted by Aladdin Two cars face each other on a horizontal road. Car A starts from rest at t=0 and travels with a constant acceleration of 60m/s^2, until it reaches a speed of 80ft/s. Afterward it maintain the same speed.Also after t=50 sec , Car B located 6000 ft down the road is traveling towards A with a constant speed of 60 ft/s. Determine the distance traveled by A when they pass each other. We've done this problem , having both cars launched at the same time. What are the changes that might occur. Thanks in advance.
Is the acceleration in m/s^2 or ft/s^2? I'll assume it's ft/s^2.

Let's make a table of information:

Car A: First let's find out where and when car A reaches 80 ft/s. (This will supposedly happen before the two cars pass each other.) Set the origin to be where the car starts and the positive direction be the direction it moves in. t will be when the car gets to that point and x will be the position of the car at that time.
t0 = 0 s......t = ?
x0 = 0 ft.....x = ?
v0 = 0 ft/s..v = 80 ft/s
a = 60 ft/s^2

We need both x and t for the second part. Let's start with x. We don't know t so we need an equation for x that doesn't include time. Thus
v^2 = v0^2 + 2a(x - x0)
This tells use that car A reaches v = 80 ft/s at a position x of 53.3333333 ft. The time that this happens at is (using any equation that we like)
v = v0 + at
gives t = 1.3333333 s.

Car A travels at this speed for 50 s - 1.3333333 s = 48.666666 s. It covers a an additional distance of (80 ft/s)(48.6666666 s) = 3893.333333 ft This puts car A at a position of x0 = 53.33333 ft + 3893.333333 ft = 3946.666666 ft at a time of 50 s.

Now we need to set up the next part...where the cars cross paths. This is going to require information from car A and car B. I will call the position where the cars pass each other X. They will do so at a time T. I will also reset the time to be 0 when car A gets to x = 3946.666666 ft.

Car A:
t0 = 0 s......................t = T
x0 = 3946.6666 ft.........x = X
v0 = 80 ft/s.................v = 80 ft/s
a = 0 ft/s^2

Car B:
t0 = 0 s............t = T
x0 = 6000 ft......x = X
v0 = -60 ft/s.....v = -60 ft/s
a = 0 ft/s^2

Can you take it from here? (Hint: Find two equations in X and T and solve them simultaneously.)

-Dan
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 Apr 2nd 2011, 11:32 AM #4 Member   Join Date: Oct 2009 Posts: 58 Thanks for the help guy's ~ Regarding post number 2. My work so far : S(A) = 0.5(6)(t^2) +0 +0 S(B) = 6000 + (-60)(t-50) + 0 Set S(A)=S(B) I get two values for t , one is accepted t=45.6sec and the other is rejected(negative sign ). Now if I assume that car A meets car B in the first region, V(A) = 0 + (6)(t) => t=13.5 sec Being 45.6 sec > 13.5 , my assumption is wrong ~ Therefore car A meets car B in the second region (Obvious) S(A) = 0.5(6)(13.5)^2 = 530.67 feet S(B) = 6000 - (60)(13.5 -50) = 8190 feet !! This is impossible , car B is heading towards Car B , so the distance between the two cars must decrease.. ! Soo ...
 Apr 2nd 2011, 01:05 PM #5 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Be sure that all the units are the same first! Then, the equation for car A is: S(A) = ut + 0.5at^2 + S That S is given by vt1 where t1 is the time elapsed after the car reached 80 ft/s S(B) is okay. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
Apr 2nd 2011, 10:35 PM   #6
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Join Date: Oct 2009
Posts: 58
All units are in the english system , so it's 6 ft/sec ...

Do you mean that for S(A) = 0.5(6)(13.3)^2 + 80(13.3) = 1594.67 feet.

Or before reaching a constant speed ?

Last edited by Aladdin; Apr 2nd 2011 at 10:46 PM.

 Apr 3rd 2011, 02:09 AM #7 Member   Join Date: Oct 2009 Posts: 58 I think I've caught the solution. Thanks for topsquark's brief explanation for his answer.
 Apr 3rd 2011, 03:36 AM #8 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Using v = u + at, you get the time taken for the car to reach 80 ft/s 80 = 0 + 6t0 t0 = 13.3 s Okay, I'm using another notation so that you are less confused. Car A will only accelerate for this time interval t0. [I'm using a = 6 ft/s^2 since you used that in your other post, which suggests that you did a typo in your first post] So, the total distance covered by car A is given by: S(A) = 0.5at^2 + S S(A) = 0.5(6)(13.3)^2 + 80t1 where t1 = t - 13.3 and t is the total time elapsed since the start of the motion of car A S(A) = 533.3 + 80t - 1067 = 80t - 533.3 t1 is the time elapsed during which the car runs at 80 ft/s, which is the net time t, minus the time already elapsed, 13.3 s. The distance travelled by car B is then: S(B) = u(t-50) = 60(t-50) = 60t - 3000 Then, S(A) + S(B) = 6000 So, you get: 80t - 533.3 + 60t - 3000 = 6000 140t = 6833.3 t = 68.1 s (3 sf) Then, put that in S(A). S(A) = 80t - 533.3 Which gives me S(A) = 4190 ft (3 sf) To check this, in 68.1 seconds, car A will cover = 0.5*(68.1 + (68.1-13.3))*80 = 4916 ft roughly and car B takes 68.1 - 50 = 18.1 seconds, which is 18.1*60 = 1086 ft Total is 1086 + 4916 = 6002 roughly ( S(A) if you take a fourth digit, you get 4194 ft which gives 6000 ft exactly) __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
 Apr 3rd 2011, 11:08 AM #9 Member   Join Date: Oct 2009 Posts: 58 Thanks Jerry, I can notice your work regarding the problem. I'm proud to belong to this site for three years Thanks for the help guy's. Last edited by Aladdin; Apr 3rd 2011 at 11:42 AM.

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