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# M2 velocity acceleration position, etc Watch

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1. i'm self teaching this and ive been completely thrown by this question on an exam paper:

A particle P moves on the x-axis. At time t seconds the velocity of P is v m s–1 in the direction of x increasing, where v is given by

v = (big swirly bracket)
8t - (3/2)t^2, 0<=t<=4
16-2t, t > 4

When t = 0, P is at the origin O.
Find
(a) the greatest speed of P in the interval 0 < t < 4,
(4)
(b) the distance of P from O when t = 4,
(3)
(c) the time at which P is instantaneously at rest for t > 4,
(1)
(d) the total distance travelled by P in the first 10 s of its motion.
(8)

i have done questions abc, and got them right (32/3, 32m, 8s are the answers) but for question d i am so confused. so far i have integrated (16-2t) and found c, then worked out that when t = 10, the position = 44.
so i would have thought this was just the answer, and the mark scheme says you have to do this first, but it then says: 'but direction changed, so t=8, s=48
hence total distance travelled = 48 +4 = 52m

what the hell? how do you know the direction has changed? why isn't it just 44? so so confused, i am freaking out about this exam

thanks, jb
2. (Original post by jumblebumble)
i'm self teaching this and ive been completely thrown by this question on an exam paper:

A particle P moves on the x-axis. At time t seconds the velocity of P is v m s–1 in the direction of x increasing, where v is given by

v = (big swirly bracket)
8t - (3/2)t^2, 0<=t<=4
16-2t, t > 4

When t = 0, P is at the origin O.
Find
(a) the greatest speed of P in the interval 0 < t < 4,
(4)
(b) the distance of P from O when t = 4,
(3)
(c) the time at which P is instantaneously at rest for t > 4,
(1)
(d) the total distance travelled by P in the first 10 s of its motion.
(8)

i have done questions abc, and got them right (32/3, 32m, 8s are the answers) but for question d i am so confused. so far i have integrated (16-2t) and found c, then worked out that when t = 10, the position = 44.
so i would have thought this was just the answer, and the mark scheme says you have to do this first, but it then says: 'but direction changed, so t=8, s=48
hence total distance travelled = 48 +4 = 52m

what the hell? how do you know the direction has changed? why isn't it just 44? so so confused, i am freaking out about this exam

thanks, jb
This would normally be very easy to miss, but the question led you through it and gave you a big clue:
Look at part c) - You found a time when the particle is at rest - i.e. it changes direction!

So basically, you were right in thinking that the total displacement is 44, but (like the mar scheme says) in order to calculate the total distance you must take into account the full path of the particle, not just its starting and finishing positions.
3. (And this is the key difference between displacement and distance, which many people never really understand!)

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Updated: December 23, 2010
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