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#1
A particle P moves in a straight line so that, at time t seconds, its acceleration a m/s^2 is given by

a=4t-t^2 0=<t=<3
a= 27/(t^2) t>3

At t = 0, P is at rest. Find the speed of P when

(a) t = 3,

(b) t = 6.

I can do the first part cos its straight forward but the second part, even when i looked at the answer and how they done it im still not sure, cos for part (b) they say that t>=3 but it says that t>3 for the second equation of a.
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#2
on the answer sheet they say t>=3 instead o it just beeing greater than 3 i dont get that specific part.
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15 years ago
#3
4.5ms^-1.
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15 years ago
#4
I got the same answer for a). For b) follow the exact same method and you should come out with 13.5m/s. I can post my workings if you like.

As to the inequalities symbols, don't worry about them to much - the difference between 'less than' and 'less than or equal to' in continuous data is essentially zero. In fact, to be strictly accurate the examiner is more correct than if they were swapped around. Again, I can post why if you're interested.
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15 years ago
#5
(Original post by lexazver203)
A particle P moves in a straight line so that, at time t seconds, its acceleration a m/s^2 is given by

a=4t-t^2 0=<t=<3
a= 27/(t^2) t>3

At t = 0, P is at rest. Find the speed of P when

(a) t = 3,

(b) t = 6.

I can do the first part cos its straight forward but the second part, even when i looked at the answer and how they done it im still not sure, cos for part (b) they say that t>=3 but it says that t>3 for the second equation of a.
a. is it 9??
b. is it 13.5??
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#6
(Original post by el GaZZa)
I got the same answer for a). For b) follow the exact same method and you should come out with 13.5m/s. I can post my workings if you like.

As to the inequalities symbols, don't worry about them to much - the difference between 'less than' and 'less than or equal to' in continuous data is essentially zero. In fact, to be strictly accurate the examiner is more correct than if they were swapped around. Again, I can post why if you're interested.

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#7
(Original post by lgs98jonee)
a. is it 9??
b. is it 13.5??
yas and yes
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15 years ago
#8
Hi wot is the confirmed answer, because origianlly i also got 13.5 ms^-2 but i have now realised i made a mistake in my working because i took t as 6 instead of 6-3 which i should have done, if this makes sense. This is what i think is teh correct working:
for the second part, take the initial speed as 9, as this is what the speed is at 3 seconds, take the time as 3, not 6 because you are modelling the particle after it has already travelled for 3 seconds by taking the initial speed as 9. put this into v = u + at
v=9 + (27/36)x3
v=9 + 27/12
v=11.25 ms^-2
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#9
(Original post by mancoolio)
Hi wot is the confirmed answer, because origianlly i also got 13.5 ms^-2 but i have now realised i made a mistake in my working because i took t as 6 instead of 6-3 which i should have done, if this makes sense. This is what i think is teh correct working:
for the second part, take the initial speed as 9, as this is what the speed is at 3 seconds, take the time as 3, not 6 because you are modelling the particle after it has already travelled for 3 seconds by taking the initial speed as 9. put this into v = u + at
v=9 + (27/36)x3
v=9 + 27/12
v=11.25 ms^-2
well the right answer is 13.5 and it is just metres per second not metres per second squared
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15 years ago
#10
(Original post by mancoolio)
Hi wot is the confirmed answer, because origianlly i also got 13.5 ms^-2 but i have now realised i made a mistake in my working because i took t as 6 instead of 6-3 which i should have done, if this makes sense. This is what i think is teh correct working:
for the second part, take the initial speed as 9, as this is what the speed is at 3 seconds, take the time as 3, not 6 because you are modelling the particle after it has already travelled for 3 seconds by taking the initial speed as 9. put this into v = u + at
v=9 + (27/36)x3
v=9 + 27/12
v=11.25 ms^-2
a = 4t - t² , 0 <= t <= 3
a = 27/t², t > 3

if you put in t=3 to both these eqns you get a = 3 both times. You use the 1st eqn to work out the accln for t up to 3 secs, and the 2nd eqn thereafter. But the movement is continuous (you haven't been told it isn't) so the accln should not suddenly change at t=3. You could even have written the eqns as,

a = 4t - t² , 0 <= t < 3
a = 27/t², t >= 3

or even,

a = 4t - t² , 0 <= t <= 3
a = 27/t², t >= 3

but that last is a bit sloppy.

All three ways of writing the eqns similarly describe the movement of the particle.

for t<=3, you get

v = 2t² - t³/3 + C1 (at t=0, v = 0, => C1 = 0)
v = 2t² - t³/3, 0 <= t <= 3

at t = 3, v = 18 - 9 = 9
v = 9 m/s
======

for t > 3,

v = -27/t + C2

and at t= 3, v = 9, regardless of which eqn of motion you use since they both (essentially) cover that time value.

Therefore C2 = 9 + 27/3
C2 = 18

v = 18 - 27/t, t > 3

at t = 6,

v = 18 - 27/6
v = 18 - 4.5
v = 13.5 m/s
========
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#11
(Original post by Fermat)
a = 4t - t² , 0 <= t <= 3
a = 27/t², t > 3

if you put in t=3 to both these eqns you get a = 3 both times. You use the 1st eqn to work out the accln for t up to 3 secs, and the 2nd eqn thereafter. But the movement is continuous (you haven't been told it isn't) so the accln should not suddenly change at t=3. You could even have written the eqns as,

a = 4t - t² , 0 <= t < 3
a = 27/t², t >= 3

or even,

a = 4t - t² , 0 <= t <= 3
a = 27/t², t >= 3

but that last is a bit sloppy.

All three ways of writing the eqns similarly describe the movement of the particle.

for t<=3, you get

v = 2t² - t³/3 + C1 (at t=0, v = 0, => C1 = 0)
v = 2t² - t³/3, 0 <= t <= 3

at t = 3, v = 18 - 9 = 9
v = 9 m/s
======

for t > 3,

v = -27/t + C2

and at t= 3, v = 9, regardless of which eqn of motion you use since they both (essentially) cover that time value.

Therefore C2 = 9 + 27/3
C2 = 18

v = 18 - 27/t, t > 3

at t = 6,

v = 18 - 27/6
v = 18 - 4.5
v = 13.5 m/s
========
thanx - very nicely explained i think i actually understand it now.
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#12
But the movement is continuous (you haven't been told it isn't) so the accln should not suddenly change at t=3. You could even have written the eqns as,

so you always assume that a stays the same unless they tell you right?
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15 years ago
#13
(Original post by lexazver203)
But the movement is continuous (you haven't been told it isn't) so the accln should not suddenly change at t=3. You could even have written the eqns as,

so you always assume that a stays the same unless they tell you right?
right 0
#14
(Original post by Fermat)
right awesome!
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15 years ago
#15
(Original post by lexazver203)
But the movement is continuous (you haven't been told it isn't) so the accln should not suddenly change at t=3. You could even have written the eqns as,

so you always assume that a stays the same unless they tell you right?
Bah, soz I had to leave unexpectedly earlier. If you want to check that the acceleration stays the same at t=3 (and it will always), stick t=3 into both equations and you will come up with the same answer (in this case acc=3).
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