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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.

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.

a. is it 9??
b. is it 13.5??

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

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
========

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

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?