# Mechanics

Watch
Announcements
Thread starter 4 weeks ago
#1
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
0
4 weeks ago
#2
(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 to get points for the sketch.
Last edited by RogerOxon; 4 weeks ago
0
Thread starter 4 weeks ago
#3
(Original post by RogerOxon)
Where did the v=40 and s=200 come from?
By using v=u+at
0
Thread starter 4 weeks ago
#4
(Original post by RogerOxon)
Where did the v=40 and s=200 come from?

I'd use to get points for the sketch.
Is there a diff method that I can use for plotting the graph?
0
4 weeks ago
#5
(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 to get points for the sketch.
You should be able to deduce its shape from the equation.
0
Thread starter 4 weeks ago
#6
(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
0
Thread starter 4 weeks ago
#7
(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 ?
0
4 weeks ago
#8
(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.
0
4 weeks ago
#9
(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.
0
Thread starter 4 weeks ago
#10
(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
0
Thread starter 4 weeks ago
#11
(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!!!
0
4 weeks ago
#12
(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.
0
Thread starter 4 weeks ago
#13
(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
0
4 weeks ago
#14
(Original post by Shas72)
Yeah true
Just for fun:

,

1
Thread starter 4 weeks ago
#15
(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?
0
4 weeks ago
#16
(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?
0
Thread starter 4 weeks ago
#17
(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
0
Thread starter 4 weeks ago
#18
(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
0
Thread starter 4 weeks ago
#19
(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
0
Thread starter 4 weeks ago
#20
(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
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Are you travelling in the Uni student travel window (3-9 Dec) to go home for Christmas?

Yes (99)
28.37%
No - I have already returned home (45)
12.89%
No - I plan on travelling outside these dates (67)
19.2%
No - I'm staying at my term time address over Christmas (37)
10.6%
No - I live at home during term anyway (101)
28.94%