Kinematics problem but no numbers given?

[jnimagine]

You drive at a constant speed of v1 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, a. Let the time at which the dragster starts to accelerate be t=0

and then I'm asked to calculate the velocity and the longest time that would take me to run into the back of the dragster after the dragster started accelerating...

um.. I'm not given any #s sooo i don't know what I'm supposed to do..

then I'm asked to calculate dragster's position at tmax, distance car travels until tmax, and starting position of the car...

How the heck do I answer these if there are no numbers given?!

but the third part then asks find numerical values for tmax and dstart... for values v1=60mph, a=50ms^2...

I am very confused...
please help me out here... what kind of answers do they want??

 

[Kurdt]

Just answer algebraically and then plug the numbers in when it asks at the end.

 

[jnimagine]

Just answer algebraically and then plug the numbers in when it asks at the end.

What I came up with was:

car = vt dragster = 1/2at^2
vt = 1/2at^2
1/2at^2 - vt = 0
but it asks us to write an equation "tmax = ... "
don't u just use quadratic equation? or am I wrong..?

and then the next question when it asks for the distance of the car from the dragster when the dragster started accelerating at t = tmax, how do I figure that out?
There are steps along the way that guide us through...
1. find the drag car position at tmax
2. distance car travels until tmax
3. starting position of car
and then finally "Dstart =..."

I am so lost...

 

[Kurdt]

Can you post the question exactly as written I'm finding it a little hard deciphering what exactly you're supposed to be doing.

The method you used in the post above is how you would find the collision time. You neglected however to take into account the cars starting position. Also what velocity is the first question referring to?

 

[jnimagine]

Can you post the question exactly as written I'm finding it a little hard deciphering what exactly you're supposed to be doing.

The method you used in the post above is how you would find the collision time. You neglected however to take into account the cars starting position. Also what velocity is the first question referring to?

sorry about that.
the background information is as i wrote it above.
The questions are as follows.
1. What is tmax, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?
a) Calculate the velocity... I'm not sure which v it's referring to...
answer: tmax = ...

2. Assuming that the dragster has started at the last instant possible ( so your front bumper almost hits the rear of the gagster at t = tmax) find your distance from the dragster when he started. If you clculate positions on the way to this solution, choose coordinates so that the position of the drag car is 0 at t=0.
a) Drag car position at time tmax
b) Distance car travels until tmax
c) starting position of car
answer: Dstart = ...

This question is worded in a very confusing way... :(

 

[jnimagine]

please help! T.T

 

[jnimagine]

anyone??

 

[Kurdt]

So you're just answering with expressions until the end when you plug in the numbers. You have the right method for finding the time as I said above, except you neglected to account for the cars initial position. one other condition that you need to be aware of if the two numerical values you are later given are to work is that when the car just misses the bumper, the dragster must be traveling at the same speed as the car at that point.

 

[jnimagine]

So you're just answering with expressions until the end when you plug in the numbers. You have the right method for finding the time as I said above, except you neglected to account for the cars initial position. one other condition that you need to be aware of if the two numerical values you are later given are to work is that when the car just misses the bumper, the dragster must be traveling at the same speed as the car at that point.

hmm... but i don't know the car's initial position... it's not given later either... then don't i just assume 0?

car = vt dragster = 1/2at^2
vt = 1/2at^2
1/2at^2 - vt = 0

If this is the correct method of finding tmax... how can you rearrage this equation so that u get "tmax = ..."

if the speed are the same... then instead of equation the distances of car and the dragster like above, do i equate the velocities of the two?? but with which equation?? because I need to know d, if I equate the velocities with the equations used above...

Does finding Dstart later have anything to do with this??

 

[jnimagine]

Would this be correct??

v of the car = v of the dragster
v of the car = v1 + at
v = at
t = v/a ??

 

[Kurdt]

You need to add an extra term to the cars position to account for the different starting position. For the condition given later I believe what you've done in post #10 is ok.

 

[jnimagine]

You need to add an extra term to the cars position to account for the different starting position. For the condition given later I believe what you've done in post #10 is ok.

i'm confused...
so for tmax = v/a... i need to add something to it?? but the starting position of the car isn't given...
for dstart,, is it just dstart = vtmax??

 

[Kurdt]

If you were to solve for t_max with the original quadratic method one would need to add an extra term. I said what you'd done in post ten was ok.

 

[jnimagine]

If you were to solve for t_max with the original quadratic method one would need to add an extra term. I said what you'd done in post ten was ok.

oh ok so i finally figured out how to get tmax! :)
then for dstart, it's just dstart = vtmax, right?

 

[Kurdt]

Not exactly. The second part says the reference for position is where the dragster starts.

 

[jnimagine]

Not exactly. The second part says the reference for position is where the dragster starts.

? so the distance of the car from the dragster when the dragster starts moving...
i don't get it...- . -

 

[Kurdt]

Well the car of course travels the same distance as the dragster, plus whatever extra distance it was away from the dragster at t=0. That extra bit is the starting position of the car.

 

[jnimagine]

Well the car of course travels the same distance as the dragster, plus whatever extra distance it was away from the dragster at t=0. That extra bit is the starting position of the car.

so I'm trying to find the distance of the car from the dragster when the dragster started to go...
but its starting position isn't given...
you're only given the car's velocity and the dragster's acceleration...

 

[Kurdt]

You have to find the starting position. They ask you to use the dragster start position as the zero point. You know however that at t max the dragster has traveled whatever distance but the car has traveled the same distance as the dragster plus the extra distance it was from the dragster at t= 0.

 

[jnimagine]

You have to find the starting position. They ask you to use the dragster start position as the zero point. You know however that at t max the dragster has traveled whatever distance but the car has traveled the same distance as the dragster plus the extra distance it was from the dragster at t= 0.

i see...
so dstart = d1 + d2
= d1 + vtmax

but how do you get d1?

 

[Kurdt]

Follow the questions instructions. What is the position of the dragster at t_max? set bot positions equal and you should be able to find the starting position of the car.

 

[jnimagine]

Follow the questions instructions. What is the position of the dragster at t_max? set bot positions equal and you should be able to find the starting position of the car.

position of the dragster would be

d = vt + 1/2at^2
= 1/2at^2

and from here, set which positions equal??

 

[Kurdt]

At the time t_max the car and the dragster will be in the same position. Set the car position at t_max equal to the dragsters, and then you can work out the initial position of the car. Just the same method you were going to use initially for finding t_max except now you're finding the starting position of the car.

 

[jnimagine]

At the time t_max the car and the dragster will be in the same position. Set the car position at t_max equal to the dragsters, and then you can work out the initial position of the car. Just the same method you were going to use initially for finding t_max except now you're finding the starting position of the car.

d/v = square root of (2d/a)
(d/v)^2 = 2d/a

and from here, i figure out d...?

 

[Kurdt]

You nearly had it.

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

Then set them equal.

 

[jnimagine]

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]

s_0 is dstart. Sorry for using different notation.

 

[jnimagine]

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]

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.

 

[jnimagine]

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]

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.

 

[jnimagine]

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]

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.

 

[jnimagine]

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]

Yes, just leave the minus sign out.