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Mechanics help

1) A particle P moves in a straight line. At time t s the displacement s cm from a
fixed point O is given by s = 1/6 ( 8 t 3 - 105 t 2 + 144 t + 540 ) . Find the distance between the points at which the particle is instantaneously at rest.
2) A particle P moves along a straight line. Initially, P is at rest at a point O on the line. At time t s, the velocity of P is v ms^(-1), where v=1/20(5-t)^(2), 0<(or equal to)t<(or equal to)8.
a) Sketch a velocity time graph for the motion of P
b) Find the values of t and the corresponding values of v when the acceleration of P is instantaneously zero.
I am struggling a lot with this and would appreciate in depth explanations with working outs to help me understand. Thankyou
Reply 1
for (1), You have an equation for the displacement of the particle at time t. Do you know how displacement and velocity are related? If a particle is at rest, what would it's velocity be?

for (2): you are given an equation for the velocity v, at time t. part a is asking you to sketch this relation, are you comfortable with sketching graphs?
for part b, its asking you a question about acceleration, do you know how acceleration and velocity are related (similar to (1))?

just getting a feel of what you are and aren't comfortable with so i can help better :smile:
Reply 2
Original post by Tayls102
for (1), You have an equation for the displacement of the particle at time t. Do you know how displacement and velocity are related? If a particle is at rest, what would it's velocity be?

for (2): you are given an equation for the velocity v, at time t. part a is asking you to sketch this relation, are you comfortable with sketching graphs?
for part b, its asking you a question about acceleration, do you know how acceleration and velocity are related (similar to (1))?

just getting a feel of what you are and aren't comfortable with so i can help better :smile:

When a particle is at rest it’s velocity would be zero and acceleration is the change in velocity per time I believe. I can sketch the graphs but I’m a bit confused with this specific one
Reply 3
okay so for (1) we need to find the points where the velocity is 0. So first we need to find an expression for velocity, velocity is the rate of change of displacement, or in other words ds/dt. Once you have found this, you can set it to 0 to find the times where the velocity is 0. Have you got any working so far which you could post?

(2) part b is similar to the above, except you want times where acceleration is 0. remember that acceleration is the rate of change of velocity, so dv/dt.


Sketching graphs:
If you follow these steps for sketching graphs it shouldn't matter what type of function you have.

1) Find the x and y intercepts, if any exist.
2) Find any stationary points of the curve, determine their location and their nature (max/min etc)
3) Determine the behaviour of the curve towards +/- infinity, and find any asymptotes of the curve (if any).

Have a go at sketching it and if you get stuck i can help.
Reply 4
Original post by Tayls102
okay so for (1) we need to find the points where the velocity is 0. So first we need to find an expression for velocity, velocity is the rate of change of displacement, or in other words ds/dt. Once you have found this, you can set it to 0 to find the times where the velocity is 0. Have you got any working so far which you could post?

(2) part b is similar to the above, except you want times where acceleration is 0. remember that acceleration is the rate of change of velocity, so dv/dt.


Sketching graphs:
If you follow these steps for sketching graphs it shouldn't matter what type of function you have.

1) Find the x and y intercepts, if any exist.
2) Find any stationary points of the curve, determine their location and their nature (max/min etc)
3) Determine the behaviour of the curve towards +/- infinity, and find any asymptotes of the curve (if any).

Have a go at sketching it and if you get stuck i can help.

image.jpg
This is how far I’ve gotten for the first one, should I just subtract the 2 distances now?
(edited 3 years ago)
Reply 5
Original post by Kffh
image.jpg
This is how far I’ve gotten for the first one, should I just subtract the 2 distances now?

yeh this looks good, yeh the difference between these distances is just the absolute value of the difference of the two distances.
One thing I would add to your working is how you got to t=8,3/4. I.e show the factorisation or the use of the quadratic formula.
Reply 6
Original post by Tayls102
yeh this looks good, yeh the difference between these distances is just the absolute value of the difference of the two distances.
One thing I would add to your working is how you got to t=8,3/4. I.e show the factorisation or the use of the quadratic formula.

image.jpg
Here is my working for q2, not sure if it’s correct
Reply 7
Original post by Kffh
image.jpg
Here is my working for q2, not sure if it’s correct

would you be able to screenshot the equation for v? from the one in the question I'm not sure if its the right one or not, v=(1/20)(5-t)2?
Reply 8
Original post by Kffh
image.jpg
Here is my working for q2, not sure if it’s correct


Original post by Tayls102
would you be able to screenshot the equation for v? from the one in the question I'm not sure if its the right one or not, v=(1/20)(5-t)2?

Here
Reply 9
Okay, double check your expanded expression for v, i think you might have made a slip up somewhere. If you post how you got to v=(1/4)t-(1/20)v3.
Reply 10
Original post by Tayls102
Okay, double check your expanded expression for v, i think you might have made a slip up somewhere. If you post how you got to v=(1/4)t-(1/20)v3.

image.jpgRealised my mistake, hope this looks better
The maths looks good, and the shape of your curve looks good to. For the graph there's a few things I'd mention:

- Remember to label both axes, i know it seems trivial but its easy marks to lose.
- Make sure to label the stationary points with both their x and y values.
- For this particular question, you only need to sketch the graph for 0<t<8. It might be worth calculating the value of the graph at t=8 and noting this on the graph just so you have the full picture.
Reply 12
Original post by Tayls102
The maths looks good, and the shape of your curve looks good to. For the graph there's a few things I'd mention:

- Remember to label both axes, i know it seems trivial but its easy marks to lose.
- Make sure to label the stationary points with both their x and y values.
- For this particular question, you only need to sketch the graph for 0<t<8. It might be worth calculating the value of the graph at t=8 and noting this on the graph just so you have the full picture.

Will do, thanks!!!
Original post by Kffh
Will do, thanks!!!

no worries, are you okay with the other parts of the question?
Reply 14
Original post by Tayls102
no worries, are you okay with the other parts of the question?

Yes thankyou
are you not supposed to use integration from the values t=8 and t=3/4?
Reply 16
Original post by DarkShadow101
are you not supposed to use integration from the values t=8 and t=3/4?

This thread is 9 months old now, but why would you need to use integration when you are given the displacement formula as part of the question?

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