# d-t graph

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#1
https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

For Q2, why isn't it V shaped???

Because it would initially decelerate and then accelerate on the way down
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7 years ago
#2
(Original post by ps1265A)
https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

For Q2, why isn't it V shaped???

Because it would initially decelerate and then accelerate on the way down
Btw this is a v-t graph, not a d-t graph.

You should get a graph which looks like a "y= - x" graph that intersects the time axis at 2 seconds. Initially velocity is maximum and it would experience a constant acceleration of -9.8 ms^-2 assuming that the positive direction is acting vertically up. Since acceleration is constant the gradient of the v-t graph should never change hence it should not be V shaped as you suggest as that would indicate that acceleration has changed.
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#3
(Original post by Jai Sandhu)
Btw this is a v-t graph, not a d-t graph.

You should get a graph which looks like a "y= - x" graph that intersects the time axis at 2 seconds. Initially velocity is maximum and it would experience a constant acceleration of -9.8 ms^-2 assuming that the positive direction is acting vertically up. Since acceleration is constant the gradient of the v-t graph should never change hence it should not be V shaped as you suggest as that would indicate that acceleration has changed.
Where does it say acceleration is constant?
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7 years ago
#4
(Original post by ps1265A)
Where does it say acceleration is constant?
It is assumed. There is only 1 force acting on the ball once it has been thrown up, we assume air resistance is negligible. This force is the gravitational force of attraction which is 9.8ms^-2 towards the centre of the earth.
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#5
(Original post by Jai Sandhu)
It is assumed. There is only 1 force acting on the ball once it has been thrown up, we assume air resistance is negligible. This force is the gravitational force of attraction which is 9.8ms^-2 towards the centre of the earth.
It's not making sense. If I throw up a ball, the initial speed and final speed are different - i.e. it's decelerating until it reaches the maximum where v= 0ms-1

Surely when the ball is coming down, the acceleration due to gravity means the its speed is increasing... this is not shown on the second part of the curve i.e. 2 seconds to 4 seconds
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7 years ago
#6
(Original post by ps1265A)
It's not making sense. If I throw up a ball, the initial speed and final speed are different - i.e. it's decelerating until it reaches the maximum where v= 0ms-1

Surely when the ball is coming down, the acceleration due to gravity means the its speed is increasing... this is not shown on the second part of the curve i.e. 2 seconds to 4 seconds
The speed is increasing in this section. The velocity is decreasing (or increasing) depending on how you have defined your directions.

Remember velocity is a vector where as speed is the modulus of velocity. So which has a greater speed a ball travelling at +1m/s or a ball travelling at -10m/s?

Clearly the second one right. It is just travelling faster in the opposite direction. The point where it intersects the x axis is at the peak of it's motion then it starts to fall down again and it's velocity increases in the opposite direction. Hence why it is shaped like it is not a v shape. Consider if it was a v shape that would mean the the velocity is never negative but it travels in 2 direction (up and down) and so since the direction changes so must the sign of the velocity. One graph in the mark scheme has defined upwards as positive the other as negative it really makes no difference as long as you are consistent. Hope this helped.

Also when Jai says the acceleration is constant he doesn't mean the ball isn't accelerating (obviously it is) he means that we are taking the acceleration due to gravity to be g=9.81 for the whole scenario. (In reality the acceleration due to gravity changes as you move distance from the surface of the earth)
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#7
(Original post by poorform)
The speed is increasing in this section. The velocity is decreasing (or increasing) depending on how you have defined your directions.

Remember velocity is a vector where as speed is the modulus of velocity. So which has a greater speed a ball travelling at +1m/s or a ball travelling at -10m/s?

Clearly the second one right. It is just travelling faster in the opposite direction. The point where it intersects the x axis is at the peak of it's motion then it starts to fall down again and it's velocity increases in the opposite direction. Hence why it is shaped like it is not a v shape. Consider if it was a v shape that would mean the the velocity is never negative but it travels in 2 direction (up and down) and so since the direction changes so must the sign of the velocity. One graph in the mark scheme has defined upwards as positive the other as negative it really makes no difference as long as you are consistent. Hope this helped.

Also when Jai says the acceleration is constant he doesn't mean the ball isn't accelerating (obviously it is) he means that we are taking the acceleration due to gravity to be g=9.81 for the whole scenario. (In reality the acceleration due to gravity changes as you move distance from the surface of the earth)
Fantastic explanation!

So if my velocity the ball is travelling downwards, and I have originally taken my direction o velocity as upwards, then my graph would go downwards as velocity in my positive direction is decreasing... But if the ball is falling downwards, the velocity is in the opposite direction to my positive direction and so the values for velocity become negative? Is that right?

So does this go for everything? Say if my diplacement is positive in the downward direction, the my velocity will also be positive?

And why was a previous question v-shaped?

https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf
Looking at Q3d), if we take the upward direction as positive, surely when the ball is going up, it's velocity is decreasing from an initial positive velocity? So why does the graph start from 0, and not this positive initial velocity?
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