# A level maths question

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#1
A particle travels between three points on a line, P, Q and R. At t=0, the particle accelerated uniformly at 3 ms^-1 for 4 seconds, until it reaches Q with a velocity of U ms^-1. It then travels at this constant velocity for 6 seconds to point R. Immediately after passing R, the particle decelerates uniformly for T seconds until comes to rest.
a) Draw a velocity time graph for the particles motion
—> done
b) Find the value of U —> 27 ms^-1
c) Given that the total displacement is 405 m, find how long after passing P the particle comes to rest
d) Find the deceleration of the particle coming to rest

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1 month ago
#2
the total displacement is the area under the velocity/time graph, if you haven't tried that, then that's maybe where you should head towards lmao gl gl
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#3
(Original post by Kei2202)
the total displacement is the area under the velocity/time graph, if you haven't tried that, then that's maybe where you should head towards lmao gl gl
The total displacement is given as 405 m.... I need to find the time taken after the particle passes point P
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1 month ago
#4
Can you post your working so far?
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1 month ago
#5
(Original post by malli07)
A particle travels between three points on a line, P, Q and R. At t=0, the particle accelerated uniformly at 3 ms^-1 for 4 seconds, until it reaches Q with a velocity of U ms^-1. It then travels at this constant velocity for 6 seconds to point R. Immediately after passing R, the particle decelerates uniformly for T seconds until comes to rest.
a) Draw a velocity time graph for the particles motion
—> done
b) Find the value of U —> 27 ms^-1
c) Given that the total displacement is 405 m, find how long after passing P the particle comes to rest
d) Find the deceleration of the particle coming to rest

Also, how did you calculate U?
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#6
(Original post by mathstutor24)
Also, how did you calculate U?
V=u+ at
= 15+3*4
=27 ms^-1
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1 month ago
#7
(Original post by malli07)
V=u+ at
= 15+3*4
=27 ms^-1
Where did the initial velocity of 15 m/s come from as it wasn't mentioned in the original question?
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#8
(Original post by mathstutor24)
Where did the initial velocity of 15 m/s come from as it wasn't mentioned in the original question?
I’ll retype the question

A particle travels between three points on a line P,Q and R. At time t=0, the particle passes P with a velocity if 15 ms^-1. Immediately after passing P, the particle accelerates uniformly at 3 ms^-2 for 4 seconds, until it reaches Q with velocity U ms^-1. It then travels at this constant velocity for 6 seconds to point R. Immediately after passing R, the particle decelerates uniformly for T seconds until it comes to rest.

a) draw a velocity time graph to show the particles motion—> done
b) Find the value of U—> 27 ms^-1
C) Given that the total displacement of the particle is 405m, find how long after passing P the particle comes to rest.
D) Find the deceleration of the particle in coming to rest
1
1 month ago
#9
Thanks!

Okay, so part c)...

Total displacement of the particle is 405m - as the particle has moved "forward" for its entire journey, displacement will be the distance between point P and when it comes to rest T seconds after it passes R. Do you know how to calculate displacement from a velocity-time graph?
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#10
(Original post by mathstutor24)
Thanks!

Okay, so part c)...

Total displacement of the particle is 405m - as the particle has moved "forward" for its entire journey, displacement will be the distance between point P and when it comes to rest T seconds after it passes R. Do you know how to calculate displacement from a velocity-time graph?
It’s the area under the graph
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1 month ago
#11
(Original post by malli07)
It’s the area under the graph
Yes - so if you know this you should be able to figure out the total time of the journey. Which part are you struggling on?
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#12
(Original post by mathstutor24)
Yes - so if you know this you should be able to figure out the total time of the journey. Which part are you struggling on?
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1 month ago
#13
(Original post by malli07)
You are told the area under the graph - use your diagram to find it using unknowns.
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1 month ago
#14
(Original post by malli07)
You don't need to use any of the suvat equations. Look at the area under the graph in terms of rectangles and triangles. Create an expression for the total area and equate this to 405. You should then be able to work out the value of T, and then the total time.
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#15
(Original post by Muttley79)
You are told the area under the graph - use your diagram to find it using unknowns.
I tried it...
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#16
(Original post by mathstutor24)
You don't need to use any of the suvat equations. Look at the area under the graph in terms of rectangles and triangles. Create an expression for the total area and equate this to 405. You should then be able to work out the value of T, and then the total time.
Ok I got it!!! Thanks.... I’m gonna go try out part D
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1 month ago
#17
(Original post by malli07)
I tried it...
Where? There is a trapezium, a rectangle and a triangle.

Trap area = 1/2(15 + 27) x 4

Now try the other bits.
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#18
(Original post by Muttley79)
Where? There is a trapezium, a rectangle and a triangle.

Trap area = 1/2(15 + 27) x 4

Now try the other bits.
I got it... thank you so much 🙂
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1 month ago
#19
(Original post by malli07)
I got it... thank you so much 🙂
You are welcome - because it asked you to draw the vel-time graph that can often be a hint that you might need to use it
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4 weeks ago
#20
(Original post by malli07)
The total displacement is given as 405 m.... I need to find the time taken after the particle passes point P
ah sorry i read the question wrong HAHAH
well you have the v/t graph up to R and you know that as from R the particle will uniformly decelerate until it's stationary. So find the displacement from P to R and you'll know the extra displacement it needs to get to 405 (which would be the area of the triangle formed between R and wherever the particle comes to rest) and since you know the initial speed at R you can work out the time it takes to decelerate. don't forget to add the extra seconds from P to R
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