# Solving Velocity Time Graph: Car 1 vs. Car 2

• Nirupt
In summary: Displacement, acceleration, initial velocity, and time?Car A moves with constant acceleration. You should know the formula for displacement in terms of time.
Nirupt

## Homework Statement

http://imageshack.us/photo/my-images/96/caracarbgraph2.png/[PLAIN]

[h2]Homework Equations[/h2]
This image is 2 cars, and Car 1 is starting at a higher velocity rate than Car 2, which is constant @ t = 0. Now then, the car is ahead and then decreases speed over time, I am trying to figure out when car 2 well eventually catch up to car 1.

[h2]The Attempt at a Solution[/h2]
I am not sure if there is any equation to this, but it looks like they go the same speed at 5 seconds, the slope of car 1 is 6 m/s^2 (70 m/s - 10 m/s = 60 m/s/10s = 6m/s^2) However I'm not even sure if that actually helps.. since car 2 has a slope of 0.

http://img96.imageshack.us/img96/7369/caracarbgraph2.png

Last edited by a moderator:
Nirupt said:

## Homework Statement

http://imageshack.us/photo/my-images/96/caracarbgraph2.png

## Homework Equations

This image is 2 cars, and Car 1 is starting at a higher velocity rate than Car 2, which is constant @ t = 0. Now then, the car is ahead and then decreases speed over time, I am trying to figure out when car 2 well eventually catch up to car 1.

## The Attempt at a Solution

I am not sure if there is any equation to this, but it looks like they go the same speed at 5 seconds, the slope of car 1 is 6 m/s^2 (70 m/s - 10 m/s = 60 m/s/10s = 6m/s^2) However I'm not even sure if that actually helps.. since car 2 has a slope of 0.

You need to know what was the initial distance between the cars.

ehild

[STRIKE]40 t = 70 t + 1/2 * (-14) t^2[/STRIKE]

40 t = 70 t + 1/2 * (-6) t^2

Last edited:
rollingstein said:
40 t = 70 t + 1/2 * (-14) t^2

That would be true when they start from the same place (if you used the correct acceleration).

ehild

ehild said:
That is true when they start from the same place.

ehild

In the absence of other info. I assumed that, yes.

Why -14?

Nirupt said:
Why -14?

What's acceleration?

That is not -14. Check.

ehild

ehild said:
That is not -14. Check.

ehild

Right. My blunder. -6.

40 t = 70 t + 1/2 * (-6) t^2

It's a quadratic right? How did you know to use that formula?
so A = -3, B = 30, and C= 0
= -5 (+/-) SQRT(1) = 0, and 10

So 10 seconds is the answer.

Nirupt said:
It's a quadratic right? How did you know to use that formula?

Which?

Nirupt said:
It's a quadratic right? How did you know to use that formula?
so A = -3, B = 30, and C= 0
= -5 (+/-) SQRT(1) = 0, and 10

So 10 seconds is the answer.

What kind of motions are involved?

ehild

Displacement, acceleration, initial velocity, and time?

Car A moves with constant acceleration. You should know the formula for displacement in terms of time.

ehild

x =v0t + ½*at2

But it was setup that x was 40t because the distance was needed between the 2 cars correct?

If the set-up was that both cars started from the same position x=0, then the position of car A is xA=70t-3t2 as function of the time t and the position of car B is xB=40t. Car A leaves car B at the start, but slows down and car B catches it at the end. At that instant, both cars are at the same place again, xA=XB. 40t=70t-3 t2

ehild

## 1. How do I read a velocity-time graph?

A velocity-time graph shows the relationship between an object's velocity and time. The y-axis represents velocity in meters per second, and the x-axis represents time in seconds. The slope of the graph represents the object's acceleration, with a steeper slope indicating a higher acceleration.

## 2. How do I determine which car has a higher velocity on a velocity-time graph?

The car with a higher velocity will have a steeper slope on the graph. This means that it is accelerating at a faster rate and covers more distance in a given amount of time.

## 3. How can I tell if two cars are traveling at the same velocity on a velocity-time graph?

If the two cars have the same velocity, their velocity-time graphs will have the same slope. This means that they are both accelerating at the same rate and covering the same distance in a given amount of time.

## 4. How can I calculate the acceleration of each car from a velocity-time graph?

The acceleration of a car can be calculated by finding the slope of the velocity-time graph. To do this, you can choose two points on the graph and use the formula acceleration = (change in velocity)/(change in time). Make sure to use consistent units for time and velocity.

## 5. Can I use a velocity-time graph to determine the distance traveled by each car?

Yes, the distance traveled by each car can be determined by finding the area under the curve on the velocity-time graph. This can be done by dividing the graph into smaller rectangles and calculating the area of each one, then adding the areas together. The units for distance will be meters, as represented on the y-axis of the graph.

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