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I am reading the tutorial on physics classroom.com, and am at Meaning of Slope for a v-t graph. It gives data for an object experiencing an acceleration:

enter image description here

This is the corresponding chart for the object's velocity:

enter image description here

At 0 seconds the position of the object is 0 meters. And the velocity is 0 m/s which makes sense. After 1 second, the object travels 5 meters. The chart says that its velocity is 10 m/s at time equals 1 second. How is this possible if the object only travelled 5 meters within that second? Shouldn't the velocity be 5 m/s and not 10 m/s?

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I fixed the formatting. Hopefully it is easier to read. –  user2620463 Feb 22 '14 at 16:07
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Related: physics.stackexchange.com/q/89590/2451 –  Qmechanic Feb 22 '14 at 16:14
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I find it odd that people are voting to close this as a homework-like question. Our homework policy was crafted to specifically allow conceptual questions like this one, so if people think this should be off topic, there are some major changes in order. –  David Z Feb 22 '14 at 22:57
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Agreed. This is a beginners question, but it seems to me it's a valid one. –  John Rennie Feb 23 '14 at 7:29

5 Answers 5

The velocity is constantly increasing due to a constant acceleration. Exactly at 1 s the velocity is 10 m/s, but this does not mean that velocity was at 10 m/s in preceding second. In fact, given the distance 5 m moved in this second, the average velocity in this second was 5 m/s. And this should make sense to you, because in this first second the velocity is linearly increasing from 0 to 10 m/s, which by average is 5 m/s.

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We can't know that the acceleration is constant. –  RdErdwien Feb 22 '14 at 17:41
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Given the velocity at timesteps from 0 to 5s, which increases every second by 10 m/s, we can easily assume a constant acceleration of 10 m/s^2. –  pressure Feb 23 '14 at 8:49

The car probably experiences a constant acceleration of $10{m/s^2}$. You can see from the chart that the velocity follows this as after every second the car is going $10{m/s^2}$ faster. However, this is clearly not the whole picture. We do not know the acceleration at 1.5 seconds, or 1.55 or 3.14.

We can get confirmation that our acceleration model works by checking it against the distances.

$\int\limits_0^1{10t\mathrm{d}t} = 5$

$\int\limits_1^2{10t\mathrm{d}t} = 15$

$\int\limits_2^3{10t\mathrm{d}t} = 25$

$\int\limits_3^4{10t\mathrm{d}t} = 35$

$\int\limits_4^5{10t\mathrm{d}t} = 45$

If you notice the integrals assuming an acceleration of 10 show the distance between each second for every distance value we have. It's a reasonable assumption to say that the acceleration is probably $10m/s^2$.

Note though that this acceleration is not the only one that could possibly produce those points. It is the simplest to find because we assumed acceleration was positive and constant. We would really need a function in order to be sure.

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I do not know integrals. I just know that they are the opposite of a derivative. So if I understand correctly, the velocity chart shows the instantaneous velocities at the given times. So at 1 second the car had an instantaneous velocity of 10 m/s. Am I correct in saying that the car did not travel 10 meters within this second, it is just travelling at that rate? If acceleration stopped, it would travel 10 meters in a second? A derivative could be used to determine the velocities. So you reversed the derivative (integration?) to get the distances. Why are they different from the chart? –  user2620463 Feb 22 '14 at 17:51
    
Yes, you are correct that the velocity chart shows the instantaneous velocities. If acceleration stopped at 1 second then the car would continue at 10m/s. It did not travel 10 m within that second. As for the integrals, they are reversed to get the distances. Notice the limits on the left side of the integrals. Those indicate seconds. Between second 0 and 1 we know the car traveled 5 meters. This is consistent with the integral assuming acceleration is 10m/s^2. That process is repeated to show that 10m/s^2 obtains the correct difference for each distance value we have. –  RdErdwien Feb 22 '14 at 17:57
    
I see now. The distances from the integrals are the change in distance within the intervals, and not the total distance travelled as shown in the diagram. –  user2620463 Feb 22 '14 at 18:16
    
So the object's velocity during the first second was an average of 5 m/s, but its instantaneous velocity was 10 m/s at the end of this interval. Could the distances from the integration be turned into average velocities? And they increase by 10 m/s each time. So 10m/s/s would just be an average acceleration? –  user2620463 Feb 22 '14 at 18:19
    
Exactly, we do see that they increase by 10 every time. They can be considered to be the average velocity in between because we are only taking one second intervals. If we were to take 2 second intervals we would have to divide by two in order to get the average velocity in those two seconds. 10m/s/s is an average acceleration, the last paragraph of my answer talks about this. Its most likely in reality a constant acceleration of 10m/s/s, but there are infinite functions that can fit our data points. –  RdErdwien Feb 22 '14 at 18:32

Let's take a look just from the point of view of someone reading the problem.

First of all, we can say "our car has motion", because it's changing its position each second.

Ok. So, how is its motion? Well, it is moving in 1 dimention, it is a linear movement. Then, we can say "our car has linear motion". Also, we can see our car's velocity is changing every second. Then, we can say "our car has linear motion, and has acceleration".

Ok. So, how is its acceleration? Well, we can see from the chart that, in this particular case, it's velocity is changing constantly. This is interesting and very important. To make a deduction from this, we need to remember how acceleration is defined:

Acceleration is defined as the rate of change of velocity with respect to time.

For those of you with knowledge of calculus, you can take this definiton as:

Acceleration is the second derivative of displacement.

But you don't need to know calculus to see that a constant change of velocity is equivalent to a constant acceleration.

Now we can say "our car has uniformly accelerated linear motion" that also means "our car has linear motion, and has a constant acceleration".

Knowing we have a constant acceleration, in order to understand what is going on, we only need 2 more things. Initial conditions, and the exact value of the acceleration.

For initial conditions you can see from the picture that at the start, time $t=0 s$, the car is where we call "0 m", or $x=0 m$. From the chart, we can read that the car has a velocity $v=0 m/s$.

So our initial conditions are: $$t=0 s$$ $$x=0 m$$ $$v=0 m/s$$

And finally, we can see each that each second the velocity increases $10 m/s$. In other words, our acceleration $a$ must be $10 m/s^2$.

Now that we have worked out the situation to the point we know everything from our problem, let's introduce your question.

What should be the velocity of the car in $t=1s$?

Going back to our definitions, if the topic of your webpage is "Meaning of Slope for a v-t graph", you must be interested in how to find out the velocity at a certain step of time.

The graph of the v-t from your car is a straight line that goes from $(t=0,v=0)$ to $(t=5,v=50)$. The slope from this graph has the same meaning as the acceleration:

The slope is defined as the rate of change of the dependent variable (velocity) with respect to the independent variable (time).

If we call $m$ the slope from the graph, by definition it should be calculated:

$$m = \frac{\Delta v}{\Delta t} = \frac{v_{j} - v_{i}}{t_{j} - t_{i}}$$

Good for any integer $i,j\leq 6$, as long as $i<j$.

For this particular case, and for any pair of data we choose, we will find:

$$m = \frac{1-0}{1-0} = \frac{10}{1} = 10$$

or, by definition of our variables:

$$a = 10 m/s^2$$.

When analyzing graphs, the area under the curve can be understood as "the integration", in this case, you can say that the area under the curve (a straight line in here) is the value of your distance, as it is the integration of your velocity through the time it has been increasing, so, as our graph draws a triangle:

$$x_{final}=\frac{(v_{i} - v_{initial})(t_{i} - t_{initial})}{2}$$

In particular:

$$x_{t=1}=\frac{(v_{t=1} - v_{initial})(1s - 0s)}{2} = \frac{(10 m/s - 0m/s)(1s)}{2}$$ $$x_{t=1}= 5 m$$

The same principle, but in a common formula that you can you without drawing the graph is:

$$x_{final}=x_{initial}+v_{initial}t+\frac{1}{2}at^2$$

$$x_{t=1}=x_{initial}+(v_{initial})(1s)+\frac{1}{2}a(1s)^2$$ $$x_{t=1}=0m+(0m/s)(1s)+\frac{1}{2}(10m/s^2)(1s^2)= \frac{1}{2}(10m)$$ $$x_{t=1}= 5 m$$

Both methods give the same result. What is messing with your thoughts are just common mistakes of interpretation of how uniformly accelerated motion works.

I hope to clear your questions.

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You must consider the acceleration that is acting on the object. Let me give you an example, a ball falls from rest to the floor. This ball is 5m above the ground and the acceleration on the ball due to gravity is 10m/s^2. When the ball is let go, clearly the ball is not moving at 0m/s which is what it would be if there were no acceleration, it would just stay in mid air. We both know that the ball gains velocity, it gets faster. The rate at which it gets faster is the acceleration. So in the first second the ball has gained velocity. This velocity is 10m/s as the acceleration is 10m/s^2. In the next second the ball gains another 10m/s so the ball travelling at two seconds the ball has a velocity of 20m/s. Continuing with this logic the ball will have a velocity of 50m/s at 5 seconds.

I hope this explanation and the one above helps you!

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This explanation did help. The ball's velocity increases by 10 m/s every second. So in the first second the ball should have a velocity of 10 m/s. Does this mean that the ball travelled a distance of 10 m, or does this mean that if acceleration stopped the ball would travel 10 m/s? –  user2620463 Feb 22 '14 at 17:39
    
The latter. Multiplying speed and time to get distance only holds when there is zero acceleration. –  EtaZetaTheta Feb 22 '14 at 20:05

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