- #1

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- Thread starter freefalling
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- #1

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- #2

- 28

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- #3

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- #4

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Do you understand what the product of a velocity-time actually is graph is?

Are you aware of the vector nature of velocity and displacement?

- #5

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Yes, but I just don't get why it's 16-2...

- #6

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Take the difference between the area above the graph and the area below the graph.

Last edited:

- #7

- 1,065

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At starting point, your reversing speed is 4m/s and you applied brake and the car slowly reducing its speed till 2 secs later your speed is zero.

At time=2 sec and your speed is zero, you start forward for 2 secs where you're back to your starting point.

And you continue moving forward at constant speed of 4m/s for 2 sec.

At t=6sec you applied brake and the car move for 2 sec and reach final speed of 2m/s.

What is your distance from the starting point?

- #8

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It doesn't say, the graph is all I got :( Thank you for your help!

- #9

- 1,065

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It doesn't say, the graph is all I got :( Thank you for your help!

That's what you should roughly say when looking and interpreting the graph. Hope it is helpful.

- #10

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So what you don't get is the reason one of the areas subtracts from the other?

- #11

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Yes :( like why it's 16-2 but not 16-6 (thats what makes sense to me)

- #12

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So, if I move forward 1m from point X, my displacement is 1m.

If I move backward 1m from point X, my displacement is -1m.

Knowing this, and knowing that we have a

If I move 1m forward and it takes me 1 second, my velocity is 1 metre per second (m/s)

If I move 1m backward and it takes me 1 second, my velocity is -1m/s.

If I move for 2 seconds at 1m/s, my displacement is 2m.

If I move for 2 seconds at -1m/s, my displacement is -2m.

Knowing all this, we can look at the graph. Clearly you know that the area under the graph gives displacement. If you don't know why:

We're multiplying the x-axis by the y-axis.

The x-axis is time and the y-axis is velocity.

Therefore, x = t, and y = d / t.

x*y = (d / t) * t = d.

d = displacement.

So looking at the graph we see that there are two distinct sections. One where velocity is negative (backwards) and one where velocity is positive (forwards). This translates into the car reversing from Point A and then moving forwards towards point A and continuing.

Clearly, because (v / t) * t = d, if we take a negative velocity and multiply it by time we'll get a negative displacement. Accordingly we will get a positive displacement from a positive velocity.

Do you understand why it is subtracted now?

- #13

- 1,065

- 10

Yes :( like why it's 16-2 but not 16-6 (thats what makes sense to me)

Velocity is a vector and time is a scalar quatity,

Displacement =V×t

Thus displacement is a vector quantity.

First 2 sec. the velocity is negative. Thus displacement is negative we call it d1

The rest is positive velocity. Positive displacement.d2.

Total displacement=(-)d1+d2.

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