# Kinematics problem but no numbers given?

You nearly had it.

$$s_d =\frac{1}{2} at^2$$
$$s_c = ut + s_0$$

Then set them equal.

oh i get it.
so i find what s0 is and then dstart = s0 + vt?

Kurdt
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s_0 is dstart. Sorry for using different notation.

s_0 is dstart. Sorry for using different notation.

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?

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.

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... - -;;

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.

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... ;;

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.

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.

Kurdt
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Yes, just leave the minus sign out.

so.... if v = 25m/s and a = 50m/s^2..
i get a negative answer...
but just write it as a positive?!

just changing the negative answer to a positive answer is allowed?!

Kurdt
Staff Emeritus
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What you're doing is finding the magnitude of a vector, except since this is 1D the magnitude will be $\sqrt{-s_0^2}$.

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?

Kurdt
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It should certainly depend on the speed of the car.

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?

Kurdt
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I don't know the system but write whatever you think it will accept.

i wrote d_start = 1/2at_max^2 - v_0t_max...
and it says it contains an incorrect multiplier or is missing one.... huh?!

Kurdt
Staff Emeritus
Gold Member
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.

so you don't think there's anything wrong with the equation, just physics wise?

Kurdt
Staff Emeritus
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No it looks fine.

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:
Kurdt
Staff Emeritus