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
Forum Admin

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,778
 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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.   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
Member

Join Date: Oct 2009
Posts: 58
 Originally Posted by Aladdin .
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.  Tags kinematics Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post nolanbradshaw01 Kinematics and Dynamics 1 Nov 20th 2015 04:38 AM amit6234623 Kinematics and Dynamics 1 Oct 8th 2015 06:13 AM hah842 Kinematics and Dynamics 2 Sep 24th 2010 08:31 AM Fazil Kinematics and Dynamics 5 Mar 20th 2009 11:47 AM devanlevin Kinematics and Dynamics 0 Nov 17th 2008 09:25 AM