Kinematics problem but no numbers given?

  • Thread starter jnimagine
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  • #26
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You nearly had it.

[tex] s_d =\frac{1}{2} at^2 [/tex]
[tex] s_c = ut + s_0[/tex]

Then set them equal.

oh i get it.
so i find what s0 is and then dstart = s0 + vt?
 
  • #27
Kurdt
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s_0 is dstart. Sorry for using different notation. :smile:
 
  • #28
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s_0 is dstart. Sorry for using different notation. :smile:

oh ok my bad..
so dstart = 1/2at^2 - vt?
but just wondering... this is when the dragster started going at the last instant possible, so, at tmax... then shouldn't it be the whole vt +d_0?? because when the dragster started to go at tmax, didn't the car cover the total distance from it going for the max. time and the starting positoin?
 
  • #29
Kurdt
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t_max is the time when the car nearly touches the dragsters bumper. The question says that the time the dragster sets off is t= 0. Thus we're setting the distances equal at t_max to find the starting distance of the car.

The car starts at some position behind the dragster when the dragster sets off (i.e. t=0). then at t_max it has covered that distance to the dragsters start position and the distance the dragster travels in a time t_max.
 
  • #30
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t_max is the time when the car nearly touches the dragsters bumper. The question says that the time the dragster sets off is t= 0. Thus we're setting the distances equal at t_max to find the starting distance of the car.

The car starts at some position behind the dragster when the dragster sets off (i.e. t=0). then at t_max it has covered that distance to the dragsters start position and the distance the dragster travels in a time t_max.

but it says "Assuming that the dragster has started at the last instant possible (so your front bumper almost hits the rear of the dragster at t = tmax), find your distance from the dragster when he started."

maybe i'm thinking too much... - -;;
 
  • #31
Kurdt
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That just means that the drag racer waits until he knows the car will just miss his bumper at the later time t = t_max. So when the dragster starts off there is a certain distance (i.e. d_start) between him and the car. If he started just before the car hit him then there would definitely be a crash unless the dragser could accelerate almost instantly.
 
  • #32
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That just means that the drag racer waits until he knows the car will just miss his bumper at the later time t = t_max. So when the dragster starts off there is a certain distance (i.e. d_start) between him and the car. If he started just before the car hit him then there would definitely be a crash unless the dragser could accelerate almost instantly.

ok... but since it's a distance, not a vector, it shouldn't have a negative sign for the answer... and when I plug in the givens to d = 1/2at^2 - v1t, I get a negative answer... ;;
 
  • #33
Kurdt
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From the way the question is set up it is in the negative x direction if you like. To make it work in the equations it must have a minus sign that is.
 
  • #34
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From the way the question is set up it is in the negative x direction if you like. To make it work in the equations it must have a minus sign that is.

so practically, i found the displacement...
a distance can't have a negative sign, right? is there a way to cancel it out somehow?? because it says in the question that the answer for the distance should never be negative.
 
  • #35
Kurdt
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Yes, just leave the minus sign out. :smile:
 
  • #36
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so.... if v = 25m/s and a = 50m/s^2..
i get a negative answer...
but just write it as a positive?!
 
  • #37
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just changing the negative answer to a positive answer is allowed?!
 
  • #38
Kurdt
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What you're doing is finding the magnitude of a vector, except since this is 1D the magnitude will be [itex] \sqrt{-s_0^2}[/itex].
 
  • #39
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ummm.. kurdt?
big problem here...
after all this work, the computer says tmax does not equal v/a....
it says that it doesn't depend on the variable, v....
what do i do!?!??!?! T.T
maybe technical error? should it be v0?
 
  • #40
Kurdt
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It should certainly depend on the speed of the car.
 
  • #41
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ok technical mistake whew!
i had to write v_0 / a
now... for dstart... since this is really sensitive about technicality...
i shoud write dstart = 1/2atmax^2 -v_0tmax right? not just t?
 
  • #42
Kurdt
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I don't know the system but write whatever you think it will accept. :smile:
 
  • #43
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i wrote d_start = 1/2at_max^2 - v_0t_max...
and it says it contains an incorrect multiplier or is missing one.... huh?!
 
  • #44
Kurdt
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I have no experience with these things. I used to hand in bits of paper with squiggles on. I suggest seeing if the thing has a FAQ or is there anyone around that you can ask for advice? A tutor maybe.
 
  • #45
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so you don't think there's anything wrong with the equation, just physics wise?
 
  • #46
Kurdt
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No it looks fine.
 
  • #47
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No it looks fine.

soo.... according to our teacher, there may be something wrong with the equation itself..T.T
not the way I input it...
he said i'm figuring out "your distance from the dragster when he started". and that i shouldn't forget the additional point
made in this question about assuming that the dragster starts at x=0 when t=0.

d_start = 1/2at_max^2 - v_0t_max

I honestly can't figure out what's wrong...
why would it say incorrect multiplier or is missing one?? grr what is wrong with this equation?!?!?!?!?!?!?
 
Last edited:
  • #48
Kurdt
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Well thats exactly what you have done and is the reason that extra distance term appears in the cars distance equation. I'm just thinking the distance equation for the car might have to subtract the the distance since it starts behind the dragster and the direction they travel in is defined positive.
 
  • #49
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Well thats exactly what you have done and is the reason that extra distance term appears in the cars distance equation. I'm just thinking the distance equation for the car might have to subtract the the distance since it starts behind the dragster and the direction they travel in is defined positive.

....
I'm still not quite sure how to change the equation to make it right..
subtract which distance?
ahhh i don't get it...!!!
 

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