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Old Mar 20th 2011, 01:42 AM   #1
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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.
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Old Mar 20th 2011, 09:56 AM   #2
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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.
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Old Mar 20th 2011, 11:21 AM   #3
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Originally Posted by Aladdin View Post
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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Old Apr 2nd 2011, 11:32 AM   #4
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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 ...
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Old Apr 2nd 2011, 01:05 PM   #5
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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.
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Old Apr 2nd 2011, 10:35 PM   #6
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Originally Posted by Aladdin View Post
.
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.
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Old Apr 3rd 2011, 02:09 AM   #7
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I think I've caught the solution. Thanks for topsquark's brief explanation for his answer.
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Old Apr 3rd 2011, 03:36 AM   #8
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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)
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Old Apr 3rd 2011, 11:08 AM   #9
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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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