A particle A moves in a straight line so that its displacement, s m, from a point O at time t s, where t is more than or equal to 0 is given by
s = t^3 - 4t^2 - 3t + 5
Find the value of t when particle is instantaneously at rest and the distance particle has then travelled (ans: 18 m)
I got 13 m... value of t is 3. Checked through my working, don't seem to detect any mistakes
Find also the value of t, to 2 d.p, when particle has returned to inital position(tried solving using long division method but can't)
adapted from 2008 add maths TYS
s = t^3 - 4t^2 - 3t + 5
ds/dt = 3t^2 - 8t - 3
When the particle is at rest ds/dt = 0
3t^2 - 8t - 3 = 0
(3t + 1) (t - 3) = 0
Therefore t = 3
Integrate [3t^2 - 8t - 3] from [0, 3] = 18m
**I think it's because you back substitute the t = 3 back into the original equation that you got 13m.
Initial position when t = 0 is s = 5.
5 = t^3 - 4t^2 - 3t + 5
0 = t (t^2 - 4t - 3)
Use roots of equation should give you
t = [4 (+-) Sqrt (28)] / 2
t = 4.645751311 or -0.645751311
Just to add on, whenever you are in doubt, draw a graph. In this case, it is 18 and not 13 because the particle was 5m away from the starting point when t=0.
And for second part, I got around 4.36s? Conflicting answers I got, so better not post. =P
Originally posted by ThunderFbolt:Just to add on, whenever you are in doubt, draw a graph. In this case, it is 18 and not 13 because the particle was 5m away from the starting point when t=0.
And for second part, I got around 4.36s? Conflicting answers I got, so better not post. =P
Deepak.C's answer is correct. Perhaps you made a careless mistake in your workings?
Originally posted by Audi:A particle A moves in a straight line so that its displacement, s m, from a point O at time t s, where t is more than or equal to 0 is given by
s = t^3 - 4t^2 - 3t + 5
Find the value of t when particle is instantaneously at rest and the distance particle has then travelled (ans: 18 m)
I got 13 m... value of t is 3. Checked through my working, don't seem to detect any mistakes
Find also the value of t, to 2 d.p, when particle has returned to inital position(tried solving using long division method but can't)
adapted from 2008 add maths TYS
TS how old in Secondary?
Originally posted by Forbiddensinner:Deepak.C's answer is correct. Perhaps you made a careless mistake in your workings?
yea, haha. got that answer with the computer's calculator, must have entered wrongly. tried with own calculator get correct answer. =)