Easy SUVAT M1 question

Question 3 Jan 2005.

http://files.physicsandmathstutor.com/download/Maths/A-level/M1/Papers-Edexcel/January%202005%20QP%20-%20M1%20Edexcel.pdf

For a) I got

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

but I'm unsure with how correct this is

Reply 1

TSRPAV

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

(edited 7 years ago)

Reply 2

TSRPAV

SUVAT comes into play later for part B) and C)

Reply 3

Mina_

Oh god hahahaha, can't believe how stupid that mistake was. Thanks!

Reply 4

Mina_

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 ..

(edited 7 years ago)

Reply 5

TSRPAV

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

Reply 6

TSRPAV

Last trapezium equation to + to 162m and =200m

And hence solve for u

Spoiler

(edited 7 years ago)

Reply 7

RDKGames

Study Forum Helper

Original post by Mina_

(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.

Reply 8

04MR17

Volunteer Section Leader

Original post by Mina_

Question 3 Jan 2005.
http://files.physicsandmathstutor.com/download/Maths/A-level/M1/Papers-Edexcel/January%202005%20QP%20-%20M1%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?

Reply 9

Mina_

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 :smile:

Reply 10

04MR17

Volunteer Section Leader

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).