Instantaneous velocity, help me please

[Veloco-raptor]

Homework Statement

I need to find the instantaneous velocity directly before a car collides with another at an intersection. It travels 24 metres with an acceleration of -5.39 m s². The only time factor given is the light sequence:
Green for 45 seconds, amber for 3 seconds and red for 5 seconds before the opposing light turns green.
I don't have the velocity or time, so I don't know how I would draw a graph to figure this out.

Homework Equations

I know the equation for Instantaneous velocity is Δy/Δx, and that I would have to draw a triangle just touching the point at which I need to find the speed of the vehicle, and then use Pythagoras. But I'm looking for speed, would instantaneous velocity still work? I have also considered the equation vi = vf + at, but again I don't have the time or velocity (neither the initial of final).

The Attempt at a Solution

If I used a displacement vs time graph, would it be the same as a velocity vs time graph? Would it give me the same (and correct) answer? Would the slope be negative, and how do I make my graph accurate enough so that my answer is?

Please, please lend me a hand, I'm so stumped.
Merci d'avance!

 

[tiny-tim]

Welcome to PF!

Hi Veloco-raptor! Welcome to PF!

I need to find the instantaneous velocity directly before a car collides with another at an intersection. It travels 24 metres with an acceleration of -5.39 m s². The only time factor given is the light sequence:
Green for 45 seconds, amber for 3 seconds and red for 5 seconds before the opposing light turns green.

I really don't understand the question … there seems a lot missing, including anything about the second car.

… I'm looking for speed, would instantaneous velocity still work?

Speed is just the magnitude of the velocity vector.

("v" is usually an abbreviation for "speed" )

If I used a displacement vs time graph, would it be the same as a velocity vs time graph?

nooo … the slope of displacement vs time is speed, the slope of speed vs time is acceleration

 

[Scintillation]

Hi,
I agree with tiny-tim. There seems to be some missing information here that is preventing us from making a correct calculation.

Because the acceleration is constant, we know that the velocity will be increasing linearly. However, this means that average velocity ≠ instantaneous velocity.

So we cannot do delta x/delta t. Another note about the graphs.
1. The slope of x vs. t is velocity. The slope of v vs. t is acceleration.
2. Similarly, the area UNDER the curve of the a vs. t graph is velocity.
And the area UNDER the curve of the v vs. t graph is displacement.

If you decide to continue in physics (I would recommend it), you will learn that these relationships are intuitively connected by the derivative (as in case one) and consequently the integral (as in case two). So as you can see, physics is fun! (no one said physics would be easy, though).

 

[Simon Bridge]

So how did you get on?
You know the acceleration and the distance ...
You need to know either the initial or final speeds, or the time involved, to draw a v-t graph. The area under the graph is the displacement and the slope of the graph is the acceleration... giving you two equations. The method is also used to complete https://www.physicsforums.com/showthread.php?p=3941528#post3941528.

You should also realize there is a limit to the kind of assistance we can give you when it's homework.

 

[asteriatic]

Dear student,

I understand that you are having trouble finding the instantaneous velocity of a car before it collides with another at an intersection. Let me provide some guidance to help you solve this problem.

Firstly, the equation for instantaneous velocity is v = Δx/Δt, where v is the velocity, Δx is the change in position, and Δt is the change in time. This means that in order to find the instantaneous velocity, you need to know the change in position and the change in time at a specific point in time.

In this case, you have been given the acceleration of the car, which is -5.39 m/s^2, and the distance it has traveled, which is 24 meters. However, you do not have the time or the velocity. In order to find the time, you can use the light sequence given in the problem. You know that the car traveled 24 meters during the green, amber, and red light phases, which lasted a total of 53 seconds. This means that the car traveled at a constant speed of 24m/53s = 0.45 m/s during this time.

Now, in order to find the instantaneous velocity just before the car collides, you will need to use the equation v = v0 + at, where v0 is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0.45 m/s, and the acceleration is -5.39 m/s^2. You can rearrange this equation to solve for t, which will give you the time just before the car collides.

Once you have the time, you can use the equation v = Δx/Δt to find the instantaneous velocity at that specific point in time. The slope of the displacement vs time graph will give you the average velocity, but in order to find the instantaneous velocity, you will need to use the equation v = Δx/Δt at a specific point in time.

I hope this helps you solve the problem. Good luck with your homework!