Mechanical principles: vehicle acceleration calculations

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STM
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Homework Statement:

Hi I am completely stuck on this part of my work, I’ve got the time graph started but am unsure what equations and how to work it out can some please help

Relevant Equations:

V=U+AT
B7DB263B-30E2-4231-9E34-CD99D3254592.jpeg
0F111859-EEBC-402D-822E-77B5CA9C9C0E.jpeg
 
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Answers and Replies

  • #2
Merlin3189
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Good start. I see you labelled your horizontal axis "t" . What is the vertical axis? (might or might not help later) You have left off the acceleration, which again might or might not help, but you have put all the other info on the graph, so might as well.

Now you need the acceleration (presumably a1, since they tell you a3 and implicitly a2)
It might or might not help to know what acceleration means?
Or you can just use a formula. In that case you'd need to find one that uses the info you have. The one you quote is ok.

Then the time for the deceleration. Exactly the same comments apply.

Then the total distance travelled. Best tackled in sections t1, t2, t3.
Here you will need to understand what velocity means, or use a different formula.
An easy start is the section t2. And that might give you a hint for the other sections.
 
  • #3
Merlin3189
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I am totally lost on this assignment as I missed a few weeks at college for personal reasons and I am really lost on this one as I didn’t cover these topics
So first, tell me what you understand by
velocity ?
acceleration?

Also, do you know any other equations for constant acceleration questions?
Have you solved any problems like this, just using the graph?
 
  • #4
Merlin3189
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So velocity is how much distance you travel per unit of time, ##v=\frac {distance \, travelled}{time \, taken}##

Acceleration is how much the velocity changes per unit of time, ##a=\frac {change\, of\, velocity}{time\, taken}##

Using that relationship;

For part (i), you know; the time taken, the velocity at the start, the velocity at the end, so you can calculate the change in velocity in the given time and then the acceleration.

For (ii) you know the acceleration, you know the velocity at the start and at the end, so you can calculate the change in velocity and the time it took.
 
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  • #5
STM
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So velocity is how much distance you travel per unit of time, ##v=\frac {distance \, travelled}{time \, taken}##

Acceleration is how much the velocity changes per unit of time, ##a=\frac {change\, of\, velocity}{time\, taken}##

Using that relationship;

For part (i), you know; the time taken, the velocity at the start, the velocity at the end, so you can calculate the change in velocity in the given time and then the acceleration.

For (ii) you know the acceleration, you know the velocity at the start and at the end, so you can calculate the change in velocity and the time it took.
12EAE78B-1CF8-4FB4-8417-91138259E866.jpeg
 
  • #6
Merlin3189
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(i) & (ii) fine.

For (iii) you need to tackle it in three sections. The formulae you are using apply to constant acceleration. But the acceleration is different in each phase.

Now, I don't know what the expressionyou quote here is supposed to do.
##\frac 1 2 a t## would give you half the change in speed over time t with acceleration a

You need to find distance.
You know all the speeds and times, so that is the sort of formula you want:
distance = function of (speeds , time)

For constant speed, distance = speed x time
But when speed is changing linearly (which is what happens when acceleration is constant) you can say
distance = (average speed) x time

I think you also need to learn (and understand) what are known as the SUVAT equations.
 
  • #7
STM
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(i) & (ii) fine.

For (iii) you need to tackle it in three sections. The formulae you are using apply to constant acceleration. But the acceleration is different in each phase.

Now, I don't know what the expressionyou quote here is supposed to do.
##\frac 1 2 a t## would give you half the change in speed over time t with acceleration a

You need to find distance.
You know all the speeds and times, so that is the sort of formula you want:
distance = function of (speeds , time)

For constant speed, distance = speed x time
But when speed is changing linearly (which is what happens when acceleration is constant) you can say
distance = (average speed) x time

I think you also need to learn (and understand) what are known as the SUVAT equations.
image.jpg
 
  • #8
STM
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On part b, I think I have the formula correct, but am I little in sure on the correct figures to put in

2155947C-9A24-4A7C-9984-14CC62FEC2BE.jpeg
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  • #9
STM
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On part b, I think I have the formula correct, but am I little in sure on the correct figures to put in

Edit ( does this look correct)

View attachment 258469
image.jpg
 
  • #10
Merlin3189
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I'm finding this a bit difficult, with the question and data in one post, then your solution in a separate one.
Also a few words would help.

I'll put the update to your previous Q, while I print out your new posts so that I can see it all in a logical sequence.
DistTime.png
 
  • #11
Merlin3189
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Yes . These answers are all near enough right.
 
  • #12
STM
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Yes . These answers are all near enough right.
Thank you for taking the time to check over my work
 
  • #13
Merlin3189
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You should note two points:

When you work out an early result, like the acceleration, you should record it to more significant figures and use that value in later calculations. If you use rounded off values in later calculations, the rounding may get multiplied and cause significant error.

Units for acceleration are metres per (second squared) ## \frac {m}{s^2} \, \text{ or } \, m/s^2##
not m/s which is speed or velocity.
 
  • #14
STM
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You should note two points:

When you work out an early result, like the acceleration, you should record it to more significant figures and use that value in later calculations. If you use rounded off values in later calculations, the rounding may get multiplied and cause significant error.

Units for acceleration are metres per (second squared) ## \frac {m}{s^2} \, \text{ or } \, m/s^2##
not m/s which is speed or velocity.

Thank you
 

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