Add. Mathematics - Integration to displacement, velocity

Audi SG

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

deepak.c SG

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

ThunderFbolt SG

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

Forbiddensinner SG

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?

marcteng SG

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?

ThunderFbolt SG

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. =)