Physics word problem- velocity, acceleration, time

AI Thread Summary
Two cars are traveling on a straight road, with one car moving at a constant velocity of 12 m/s and the other approaching at 25 m/s before applying brakes with a constant acceleration of -0.20 m/s². The discussion revolves around determining if and when the cars will crash, starting from a 200m distance apart. Calculations suggest that the cars will collide approximately 13 seconds after the second car begins braking, covering a distance of about 346 meters. The conversation highlights the importance of setting up the correct equations to solve for time and distance in physics problems.
meredith
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Homework Statement



two cars are traveling along a straight road, one behind the other. the first is traveling at a constant velocity of 12 m/s. the second, approaching from the rear, is traveling at 25 m/s. when the second car is 200m behind the first, the driver applies the brakes, producing a constant acceleration of -0.20 m/s^2. will the cars crash and if so, where and when?


Homework Equations


acceleration = (change in velocity)/(change in time)
change in dispacement = (Initial velocity) (time) + (1/2) (acceleration) (time sqaured)
velocity = (acceleration) (time)


The Attempt at a Solution


i got they woud crash after 50 meters? but i don't think that's right :(
 
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Welcome to PF!

Hi meredith ! Welcome to PF! :smile:

Show us what you did, and then we can see where it went wrong. :smile:
 
If the second car had a velocity of 12m/s^2, then they wouldn't crash, so we try to see if they do;

25 - 0,2t = 12 => t=65s

in 65 seconds we see that the cars actually crash. So its time to see where they actually crash..

so in "t" seconds after the break, the second cars velocity will be "25-0,2t" and first velocity was "25". Time past is "t" and the length passed is; "12t+200m". So if we equalize these, we have;


(25+25-0,2t)*t/2=12t+200

then we find t=13(approximately) (At the 13th second they crash) so then we have a velocity of "22,4m/s^2" of the second car when they crash and this hapens after "346" meters passed...

I hope I am correct!
 
MrEnergy said:
(25+25-0,2t)*t/2=12t+200

Yes … but how did you get "t=13(approximately)" ?

This is a quadratic equation … find the exact solution! :smile:
 
I think you're wrong.
First I think we need to set an equality of the distance, having 25t-.2t2+200=12t.
So we solve for t, getting t1=-12.85s, and t2=77.85s, so t2 is out real answer.
Then you plug t in a distance formula. So based on the first car you have d = 12m/s(77.85s)=934.2m.
 
Sakha, after 65 seconds second car's velocity becomes 12m/s^2, and they are not supposed to crash with same velocities. So that your t2=77 second is impossible...

The right answer is 13 seconds(I could'nt solve but you can use a program to get the exact number). After 13 secs they crash...
 
MrEnergy said:
… I could'nt solve …

Yes you can! :rolleyes:

It's 0.1 t2 - 13t + 200 = 0, so the answer is … ? :smile:
 
I mean't i didn't work for it:P Just used a program.
 


i just registered for this site today and i was wondering how i go about asking questions?
 
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Welcome to PF!

rmax58 said:
i just registered for this site today and i was wondering how i go about asking questions?

Hi rmax58! Welcome to PF! :smile:

Go to the https://www.physicsforums.com/forumdisplay.php?f=152" forum, choose the appropriate sub-forum, and click the "NEW TOPIC" button. :smile:

(and read the homework FAQs! :wink:)
 
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