Shas72
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A firework takes off from ground level, accelerating upwards for 10s with constant acceleration 4m/s^2.
Sketch the displacement time graph.
So I did u= 0, so v= u+at, v= 40m/s, s=vt-1/2at^2, s=200m. Is this correct way of doing the calculation?
Last edited by Shas72; 4 weeks ago
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RogerOxon
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(Original post by Shas72)
A firework takes off from ground level, accelerating upwards for 10s with constant acceleration 4m/s^2.
Sketch the displacement time graph.
So I did u= 0, so v= u+at, v= 40m/s, s=vt-1/2at^2, s=200m. Is this correct way of doing the calculation?
Where did the v=40 and s=200 come from?

I'd use s=ut+\frac{at^2}{2} to get points for the sketch.
Last edited by RogerOxon; 4 weeks ago
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Shas72
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(Original post by RogerOxon)
Where did the v=40 and s=200 come from?
By using v=u+at
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Shas72
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(Original post by RogerOxon)
Where did the v=40 and s=200 come from?

I'd use s=ut+\frac{at^2}{2} to get points for the sketch.
Is there a diff method that I can use for plotting the graph?
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RogerOxon
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(Original post by Shas72)
By using v=u+at
You don't need to calculate v before s. As per my edit above:

(Original post by RogerOxon)
I'd use s=ut+\frac{at^2}{2} to get points for the sketch.
You should be able to deduce its shape from the equation.
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Shas72
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(Original post by RogerOxon)
You don't need to calculate v before s. As per my edit above:


You should be able to deduce its shape from the equation.
Iam not having problems in plotting or sketching the graph. My problem is I'm not able to understand what formulas to use to get to the values
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Shas72
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(Original post by RogerOxon)
You don't need to calculate v before s. As per my edit above:


You should be able to deduce its shape from the equation.
So is it that we know u=0 in this case, t=10s and A= 4m/s^2, so we make use of the formula that has these ?
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RogerOxon
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(Original post by Shas72)
Iam not having problems in plotting or sketching the graph. My problem is I'm not able to understand what formulas to use to get to the values
You are asked to sketch a time-displacement graph, i.e. t and s. You know that u=0 (t=0). You want the simplest formula for s, when you know u, a and t.
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RogerOxon
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(Original post by Shas72)
So is it that we know u=0 in this case, t=10s and A= 4m/s^2, so we make use of the formula that has these ?
Yes. Look through the equations that you have, and see which is the most direct for what you want. When u=0, that simplifies many of them.
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Shas72
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(Original post by RogerOxon)
You are asked to sketch a time-displacement graph, i.e. t and s. You know that u=0 (t=0). You want the simplest formula for s, when you know u, a and t.
Thanks that was very helpful. I was doing all unnecessary stuff of finding v
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Shas72
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(Original post by RogerOxon)
Yes. Look through the equations that you have, and see which is the most direct for what you want. When u=0, that simplifies many of them.
Yeah you are right!!! Thanks!!!
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RogerOxon
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(Original post by Shas72)
Thanks that was very helpful. I was doing all unnecessary stuff of finding v
Good. It should simplify to the same forumula, but will add extra steps, with the associated possibility of mistakes.
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Shas72
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(Original post by RogerOxon)
Good. It should simplify to the same forumula, but will add extra steps, with the associated possibility of mistakes.
Yeah true
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RogerOxon
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(Original post by Shas72)
Yeah true
Just for fun:

s=vt-\frac{at^2}{2}, v=u+at

\therefore s=(u+at)t-\frac{at^2}{2}=ut+at^2-\frac{at^2}{2}=ut+\frac{at^2}{2}
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Shas72
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(Original post by RogerOxon)
Good. It should simplify to the same forumula, but will add extra steps, with the associated possibility of mistakes.
Can you pls advise on this one .
A ball is thrown upwards from a point 1m above the ground with initial speed 5m/s. It accelerates downwards at a constant rate of 10m/s^2 until it stops moving upwards, when it is caught by someone standing on a ladder.
Sketch the graph
According to what you have advised, s=1m, u=5m/s, a=10m/s^2, v=0 as it stops.
Should I use the same formula?
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RogerOxon
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(Original post by Shas72)
Can you pls advise on this one .
A ball is thrown upwards from a point 1m above the ground with initial speed 5m/s. It accelerates downwards at a constant rate of 10m/s^2 until it stops moving upwards, when it is caught by someone standing on a ladder.
Sketch the graph
According to what you have advised, s=1m, u=5m/s, a=10m/s^2, v=0 as it stops.
Should I use the same formula?
Be careful with your signs. The acceleration and initial speed are in different directions.

You also need to be careful about where you're measuring s from. All SUVAT equations implicitly have s=0 at t=0.

There are different ways of approaching this - it's all about the one that is simplest or makes the most sense to you.

As your goal is to sketch the graph, and you know that the ball is caught when v=0 (vertically), you can quickly get the time for that. Can you see how?

I would use that time to calculate s. Which SUVAT equation (knowing u, v=0, a and t) is the simplest?
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Shas72
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(Original post by RogerOxon)
Be careful with your signs. The acceleration and initial speed are in different directions.

You also need to be careful about where you're measuring s from. All SUVAT equations implicitly have s=0 at t=0.

There are different ways of approaching this - it's all about the one that is simplest or makes the most sense to you.

As your goal is to sketch the graph, and you know that the ball is caught when v=0 (vertically), you can quickly get the time for that. Can you see how?

I would use that time to calculate s. Which SUVAT equation (knowing u, v=0, a and t) is the simplest?
Is it v=u+at
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Shas72
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(Original post by RogerOxon)
Be careful with your signs. The acceleration and initial speed are in different directions.

You also need to be careful about where you're measuring s from. All SUVAT equations implicitly have s=0 at t=0.

There are different ways of approaching this - it's all about the one that is simplest or makes the most sense to you.

As your goal is to sketch the graph, and you know that the ball is caught when v=0 (vertically), you can quickly get the time for that. Can you see how?

I would use that time to calculate s. Which SUVAT equation (knowing u, v=0, a and t) is the simplest?
So t=0.5s
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Shas72
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(Original post by RogerOxon)
Be careful with your signs. The acceleration and initial speed are in different directions.

You also need to be careful about where you're measuring s from. All SUVAT equations implicitly have s=0 at t=0.

There are different ways of approaching this - it's all about the one that is simplest or makes the most sense to you.

As your goal is to sketch the graph, and you know that the ball is caught when v=0 (vertically), you can quickly get the time for that. Can you see how?

I would use that time to calculate s. Which SUVAT equation (knowing u, v=0, a and t) is the simplest?
So using s= vt-1/2at^2
I get s=1.25m
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Shas72
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(Original post by RogerOxon)
Be careful with your signs. The acceleration and initial speed are in different directions.

You also need to be careful about where you're measuring s from. All SUVAT equations implicitly have s=0 at t=0.

There are different ways of approaching this - it's all about the one that is simplest or makes the most sense to you.

As your goal is to sketch the graph, and you know that the ball is caught when v=0 (vertically), you can quickly get the time for that. Can you see how?

I would use that time to calculate s. Which SUVAT equation (knowing u, v=0, a and t) is the simplest?
Total distance would be 1+ 1.25 =2.25m
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