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# Easy SUVAT M1 question watch

1. Question 3 Jan 2005.

http://files.physicsandmathstutor.co...%20Edexcel.pdf

For a) I got

S = ?
U = 0
V = 9
A = ?
T = 20

but I'm unsure with how correct this is
2. For a)

used areas as the distance is the area under the speed-time graph as Speed=Distance/Time so Distance=Speed*Time so do the area of the right angle triangle plus the area of the square upto that time. Make sure you take length of square as 16 and not 20!! Look at the diagram carefully
3. SUVAT comes into play later for part B) and C)
4. Oh god hahahaha, can't believe how stupid that mistake was. Thanks!
5. For part b) the distance between 20s - 25s would be 38m right? Because 200m - 162m = 38m
and the area accounts for distance but if you work out the area for the trapezium it would be (4+9 /2 ) * 5 which is 32.5 ..
6. Yes for a) it's 162m and yes you could do that for b) but I would try to calculate an area=200 so you know that the triangle and square make 162m,then make an equation with u for the last trapezium and equate it to 200m
7. Last trapezium equation to + to 162m and =200m

And hence solve for u
Spoiler:
Show
Trapezium equation =1/2(parallel a+ longer longer parallel side b)*hieght =1/2(9+u)*5
8. (Original post by Mina_)
Question 3 Jan 2005.

http://files.physicsandmathstutor.co...%20Edexcel.pdf

For a) I got

S = ?
U = 0
V = 9
A = ?
T = 20

but I'm unsure with how correct this is
(a) It's asking for distance covered which would equate to the area. If you consider the first 20 seconds you can split it into a triangle with length of time 4 seconds and the a rectangle with length of time 16 seconds. Add these areas together.

(b) The last part of the graph is a trapezium. Area of that trapezium and the answer from part (a) should make up 200 in terms of u. Solve for u.

(c) Deceleration would be the gradient of the line between 20 and 25 seconds.
9. (Original post by Mina_)
Question 3 Jan 2005.
http://files.physicsandmathstutor.co...%20Edexcel.pdf
For a) I got
S = ?
U = 0
V = 9
A = ?
T = 20
but I'm unsure with how correct this is
In a velocity-time graph (or speed-time graph) distance is represented by the area under the line.

What I'm wondering is why this question now? Are you learning M1 over the summer?
10. (Original post by 04MR17)
In a velocity-time graph (or speed-time graph) distance is represented by the area under the line.

What I'm wondering is why this question now? Are you learning M1 over the summer?
We've finished kinematics right before summer term started so right now I'm consolidating
11. (Original post by Mina_)
We've finished kinematics right before summer term started so right now I'm consolidating
Nice. I'm learning it over the summer. Purely because I don't want to be struggling with it when the physics people are whinging about how easy it is and how they don't understand how it can be hard. I've done vectors and am now halfway through kinematics (watching examsolutions videos).

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