AQA M1 Mechanics 1B kinematics in one dimension question

Right, i do edexcel maths but i still know how to answer this!
I think perhaps drawing a velocity time/ distance graph could help. To visualise t, and create a formula. (Im speaking crap right now, but thinking lol)
I feel like you are going to calculate the area underneath your graph, distance travelled. Axes are velocity on y, time on x. Use the trapezium formula

I feel like id write something like...
1.5T +v(30-T) =313

(edited 8 years ago)

Reply 2

Don't worry about the trapezium rule, but yes, a diagram and the distances are key here.

Some ideas for your diagram: If it's accelerating for T seconds out of 30, how long is it going at constant velocity for?

What's the distance travelled while it's accelerating in terms of T? How about for the second part of the journey?

Reply 3

OP

Original post by SeanFM

Don't worry about the trapezium rule, but yes, a diagram and the distances are key here.

Some ideas for your diagram: If it's accelerating for T seconds out of 30, how long is it going at constant velocity for?

What's the distance travelled while it's accelerating in terms of T? How about for the second part of the journey?

the only thing I know to put on a velocity time graph is 30 seconds, there isn't enough data for me to do it that way

Reply 4

Original post by Sayless

the only thing I know to put on a velocity time graph is 30 seconds, there isn't enough data for me to do it that way

There is. An acceleration of 1.5 m/s^2 means that every second the velocity has increased by 1.5 m/s (try it with v = u+at) and that goes on for T seconds.. and then whatever velocity it ends up at, it stays that way until the end of the time.

Reply 5

OP

Original post by SeanFM

There is. An acceleration of 1.5 m/s^2 means that every second the velocity has increased by 1.5 m/s (try it with v = u+at) and that goes on for T seconds.. and then whatever velocity it ends up at, it stays that way until the end of the time.

I got v = 1.5T
then i used s= 1/2(u+v)t
I got 312 = 1/2(1.5T+1.5T) x (30-T)
312 = (1.5t)(30-t)
312 = 45T - 1.5T^2
1.5T^2 - 45T - 312 = 0
T^2 - 30T - 208 = 0
I don't know what to do next I can't factorise this

Reply 6

Original post by Sayless

I'm stuck on the following question 19a)

I have to find how long the car is accelerating for however I have two unknowns whenever I try make a suvat equation, I only know
a = 1.5
u = 0
I don't know anything else for that time interval
The distance i'm giving is overall, not that for time period, and the time is overall also which doesn't help, how can I answer this question?

s=ut+\frac{at^2}{2}

Using that equation, the distance the car accelerates over is

s=\frac{1.5T^2}{2}

speed = distance/time

Using that, you can find

v=\frac{312-s}{30-T}

t=\frac{v-u}{a}

So:

30=\frac{(\frac{312-(\frac{1.5T^2}{2})}{30-T})}{1.5}

That's my thinking

EDIT: with the correction it should simplify like so:

45=\frac{312-0.75T^2}{30-T}

meaning

45(30-T)=312-0.75T^2

Expanding

1350-45T=312-0.75T^2

Rearranging

\frac{3T^2}{4} - 45T + 1038=0

Getting rid of the fraction

12T^2 - 180T + 4152 =0

12(T^2 - 15T + 346) =0

T^2 - 15T+346=0

Is that the same quadratic?

(edited 8 years ago)

Reply 7

Original post by Sayless

I got v = 1.5T
then i used s= 1/2(u+v)t
I got 312 = 1/2(1.5T+1.5T) x (30-T)
312 = (1.5t)(30-t)
312 = 45T - 1.5T^2
1.5T^2 - 45T - 312 = 0
T^2 - 30T - 208 = 0
I don't know what to do next I can't factorise this

Quadratic formula maybe?

Reply 8

312 is not equal to what you've written there, though your calculation is correct for part of the distance.

The vehicle also covers some distance while it is accelerating, and from your diagram you can see that it's a neat triangle, and use what you know about the area of a velocity-time graph.

Original post by Sayless

I got v = 1.5T
then i used s= 1/2(u+v)t
I got 312 = 1/2(1.5T+1.5T) x (30-T)
312 = (1.5t)(30-t)
312 = 45T - 1.5T^2
1.5T^2 - 45T - 312 = 0
T^2 - 30T - 208 = 0
I don't know what to do next I can't factorise this

Reply 9

OP

Original post by SeanFM

312 is not equal to what you've written there, though your calculation is correct for part of the distance.

The vehicle also covers some distance while it is accelerating, and from your diagram you can see that it's a neat triangle, and use what you know about the area of a velocity-time graph.

okay i did 1.5T^2 / 2 +1.5T(30-T)=312
got T^2 - 60 + 208 = 0
got T = 56.3 or T = 3.69 which are both wrong
i have no idea what i'm doing

Reply 10

Original post by SeanFM

312 is not equal to what you've written there, though your calculation is correct for part of the distance.

The vehicle also covers some distance while it is accelerating, and from your diagram you can see that it's a neat triangle, and use what you know about the area of a velocity-time graph.

Out of curiosity, would my method work?

Reply 11

Original post by Sayless

okay i did 1.5T^2 / 2 +1.5T(30-T)=312
got T^2 - 60 + 208 = 0
got T = 56.3 or T = 3.69 which are both wrong
i have no idea what i'm doing

In your working you have skipped a few steps, but when you multiplied everything by 4/3 you got from 0.75T^2 to 1T^2, -45T to -60T (both increases) but 312 to 208 (decrease).

But the important thing is do you understand the method you're using and what it's finding out?

Reply 12

OP

Original post by SeanFM

In your working you have skipped a few steps, but when you multiplied everything by 4/3 you got from 0.75T^2 to 1T^2, -45T to -60T (both increases) but 312 to 208 (decrease).

But the important thing is do you understand the method you're using and what it's finding out?

got the right answer now
yeah i think so, im basically using v= 1.5T and using the time in terms of T and 30 seconds on the velocity time graph to find the area of the graph to work out the distnace,
so 1.5T^2 / 2 is distance of accelerating and 1.5T(30-T) is the distance of constant speed and the total distance is 312 so make it all = to 312 then make it = to 0, then make a quadratic equation to get 2 values, then eliminate the one higher than 30
thanks! I think I got it now

Reply 13

Original post by Andy98

s=ut+\frac{at^2}{2}

Using that equation, the distance the car accelerates over is

s=\frac{1.5T^2}{2}

speed = distance/time

Using that, you can find

v=\frac{312-s}{30}

t=\frac{v-u}{a}

So:

30=\frac{(\frac{312-(\frac{1.5T^2}{2})}{30})}{1.5}

That's my thinking

For the 'v' bit from the numerator you're calculating the distance travelled when the velocity is constant, but this happens for (30-T) seconds rather than 30.

As for what follows I can't fully follow what you're doing, like what the t is or how you get the equation after that. But if you get the correct quadratic you've done it right. :smile:

Reply 14

Original post by Sayless

got the right answer now
yeah i think so, im basically using v= 1.5T and using the time in terms of T and 30 seconds on the velocity time graph to find the area of the graph to work out the distnace,
so 1.5T^2 / 2 is distance of accelerating and 1.5T(30-T) is the distance of constant speed and the total distance is 312 so make it all = to 312 then make it = to 0, then make a quadratic equation to get 2 values, then eliminate the one higher than 30
thanks! I think I got it now

Well done

Reply 15

Original post by SeanFM

For the 'v' bit from the numerator you're calculating the distance travelled when the velocity is constant, but this happens for (30-T) seconds rather than 30.

As for what follows I can't fully follow what you're doing, like what the t is or how you get the equation after that. But if you get the correct quadratic you've done it right. :smile:

Oops

Reply 16

Marxist

10

Original post by Sayless

got the right answer now
yeah i think so, im basically using v= 1.5T and using the time in terms of T and 30 seconds on the velocity time graph to find the area of the graph to work out the distnace,
so 1.5T^2 / 2 is distance of accelerating and 1.5T(30-T) is the distance of constant speed and the total distance is 312 so make it all = to 312 then make it = to 0, then make a quadratic equation to get 2 values, then eliminate the one higher than 30
thanks! I think I got it now

Yup. That's it.

(edited 8 years ago)

Reply 17

Marxist

10

Original post by Andy98

s=ut+\frac{at^2}{2}

Using that equation, the distance the car accelerates over is

s=\frac{1.5T^2}{2}

speed = distance/time

Using that, you can find

v=\frac{312-s}{30-T}

t=\frac{v-u}{a}

So:

30=\frac{(\frac{312-(\frac{1.5T^2}{2})}{30-T})}{1.5}

That's my thinking

EDIT: with the correction it should simplify like so:

45=\frac{312-0.75T^2}{30-T}

meaning

45(30-T)=312-0.75T^2

Expanding

1350-45T=312-0.75T^2

Rearranging

\frac{3T^2}{4} - 45T + 1038=0

Getting rid of the fraction

12T^2 - 180T + 4152 =0

12(T^2 - 15T + 346) =0

T^2 - 15T+346=0

Is that the same quadratic?

It's definitely not that quadratic.

Reply 18