Maths M1 - Easy Mechanics question!

June 2009 - OCR (not mei) 4728 - Question 2 (!!!)

The driver of a car accelerating uniformly from rest sees an obstruction. She brakes immediately bringing the car to rest with constant deceleration at a distance of 6m from its starting point. The car travels in a straight line and is in motion for 3 seconds.

i) Sketch the (t, v) graph for the cars motion (2 marks)
Easy enough! but just to check, I always draw a (v, t) graph with time on the x- axis in questions like this. Why does it always ask for a (t, v) graph?

**** ii) *****
Calculate the maximum speed of the car during its motion (3 marks)
It's only question 2, this should be really easy but I can't seem to get my head around it! We cannot assume the time accelerating to be the same as the time decelerating, as it does not state the acceleration/ deceleration is at the same rate. Any ideas?!! / similar examples

iii) Want to give this question a shot first before asking about it. It depends on the answer from the previous part. (4 marks)

TOTAL: 9 MARKS
Thanks! :smile:

Reply 1

RVNmax

19

Original post by AidanBW

June 2009 - OCR (not mei) 4728 - Question 2 (!!!)

The driver of a car accelerating uniformly from rest sees an obstruction. She brakes immediately bringing the car to rest with constant deceleration at a distance of 6m from its starting point. The car travels in a straight line and is in motion for 3 seconds.

i) Sketch the (t, v) graph for the cars motion (2 marks)
Easy enough! but just to check, I always draw a (v, t) graph with time on the x- axis in questions like this. Why does it always ask for a (t, v) graph?

**** ii) *****
Calculate the maximum speed of the car during its motion (3 marks)
It's only question 2, this should be really easy but I can't seem to get my head around it! We cannot assume the time accelerating to be the same as the time decelerating, as it does not state the acceleration/ deceleration is at the same rate. Any ideas?!! / similar examples

iii) Want to give this question a shot first before asking about it. It depends on the answer from the previous part. (4 marks)

TOTAL: 9 MARKS
Thanks! :smile:

for i) does the word 'co-ordinate' give you a bit of a clue?
(Sorry, trying not to tell you the answer directly) :colondollar:

I always draw a (v, t) graph

I don't think you really meant that bit then - what is a (v, t) graph supposed to be? :wink:

ii) First of all, can you describe to me, from what you gather from the description, at what point do you think the car had maximum speed. It should be easy to gather from from reading, so try that first.
To check, you should get the same value by reading off the graph that you drew for part i), no?

(edited 10 years ago)

Reply 2

username1258398

19

Original post by AidanBW

i) Sketch the (t, v) graph for the cars motion (2 marks)
Easy enough! but just to check, I always draw a (v, t) graph with time on the x- axis in questions like this. Why does it always ask for a (t, v) graph?

Co-ordinates are in the form (x, y). Thus for (t, v), t is x, v is y. A "(v, t) graph" would have time on the y-axis and velocity on the x-axis.

Reply 3

AidanBW

OP

Hi ! Thanks so much both for your answers !

i) Co-ordinates! of course I see now! Thanks ! :biggrin:

(x axis - time , y axis - velocity) therefore a (t, v) graph! :smile:

ii) Maximum speed = highest point (velocity) on graph, as graph does not go negative. Therefore, maximum speed is going to be at the top of the triangle (upside-down V as written on mark scheme), although I stand by that we cannot assume each line has the same gradient.

Acceleration and deceleration values are not given - what would be the gradient of each line.
Area beneath the two lines = 6m
Time = 3 secs @ the bottom of the second line when v=0 again

I've labelled the point @ maximum speed Vmax on my graph. We must calculate this value (?)

Reply 4

AidanBW

OP

Original post by StrangeBanana

Co-ordinates are in the form (x, y). Thus for (t, v), t is x, v is y. A "(v, t) graph" would have time on the y-axis and velocity on the x-axis.

Spoilers haha! :eek:

Only just read your message though - thanks! :tongue:

Reply 5

RVNmax

19

Original post by AidanBW

Hi ! Thanks so much both for your answers !

i) Co-ordinates! of course I see now! Thanks ! :biggrin:

(x axis - time , y axis - velocity) therefore a (t, v) graph! :smile:

ii) Maximum speed = highest point (velocity) on graph, as graph does not go negative. Therefore, maximum speed is going to be at the top of the triangle (upside-down V as written on mark scheme), although I stand by that we cannot assume each line has the same gradient.

Acceleration and deceleration values are not given - what would be the gradient of each line.
Area beneath the two lines = 6m
Time = 3 secs @ the bottom of the second line when v=0 again

I've labelled the point @ maximum speed Vmax on my graph. We must calculate this value (?)

It would help if you know SUVAT equations (Even better if can understand the intuition behind them). There is one equation that you could use as you have 3 variables given, and require one to find.

Area beneath the two lines = 6m

this statement is not correct.

Unfortunately, they could have worded it more clearly in my opinion. By 6m from starting point, it is unlikely that they meant the starting point from when the car was already accelerating straight away. I'm pretty sure they meant starting point of the what was happening in that sentence. (They should really say what starting point they are referring to.)

Reply 6

AidanBW

OP

Original post by RVNmax

It would help if you know SUVAT equations (Even better if can understand the intuition behind them). There is one equation that you could use as you have 3 variables given, and require one to find.

this statement is not correct.

Unfortunately, they could have worded it more clearly in my opinion. By 6m from starting point, it is unlikely that they meant the starting point from when the car was already accelerating straight away. I'm pretty sure they meant starting point of the what was happening in that sentence. (They should really say what starting point they are referring to.)

Oh!! That makes so much more sense. Thanks! Yes- totally agree, I was thinking the whole time 6m from when the car started accelerating, that wouldn't provide enough information to use SUVAT - but if we take it as from when the car starts to decelerate (?) maybe that'll work! Thanks :smile:

Reply 7

RVNmax

19

Original post by AidanBW

Oh!! That makes so much more sense. Thanks! Yes- totally agree, I was thinking the whole time 6m from when the car started accelerating, that wouldn't provide enough information to use SUVAT - but if we take it as from when the car starts to decelerate (?) maybe that'll work! Thanks :smile:

Yep, that's exactly what I was getting at. Are you okay with that then?