Question

The velocity function is v(t)= -t^2+3t-2 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [0,5].

Answer 1

well since they want displacement (vector) not distance (scalar) watch out for sign reversals. Like when is v = 0 ???
t^2 - 3t +2 = 0
(t -1)(t-2) = 0
so do three intervals
(a) t = 0 to t = 1
(b) t = 1 to t = 2
(c) t = 2 to t = 5

Answer 2

integral from p to q of (-t^2+3t-2 ) dt =-t^3/3 + 3t^2/2 - 2 t at q - at p
= (-q^3+p^3) + (3/2)(q^2-p^2) -2(q-p)
(a) from p=0 to q=1
= -1 + 3/2 - 2 = -3 + 3/2 = -3/2
(b) etc etc etc

Answer 3

I am still getting it wrong

Answer 4

since they want displacement, ignore sign changes.
∫[0,5] -t^2+3t-2 dt = -86/5

Answer 5

-86/5 isn't right

Answer 6

in that case there is a mistake in the problem or in the answer key. All it wants is just a straight integral. You can see that the particle moves
left for 1 second
right for 1 second
left for 3 more seconds. It will wind up to the left of where it started.

Answer 7

displacement has to include sign changes
if you start at home, walk north a mile, then south a mile
your distance is TWO miles
your displacement is ZERO miles

Answer 8

its asking for total distance traveled so the answer wont be negative

Answer 9

Yes, there must be sign changes

Answer 10

It asked for the NET distance, not the TOTAL distance
Yes total is to the left

Answer 11

Huh? It said
Find the displacement (net distance covered)
not distance traveled.
But, in that case, you want

-∫[0,1] (-t^2+3t-2) dt + ∫[1,2] (-t^2+3t-2) dt - ∫[2,5] (-t^2+3t-2) dt
= 5/6 + 1/6 + 27/2
= 29/2

Answer 12

answer key said incorrect to 29/2

Answer 13

or that could be just ∫[0,5] |-t^2+3t-2| dt = 29/2
but to do the actual evaluation, you need to split it up as above.

Answer 14

well, something is wrong. We've done it both ways.
Maybe the answer key wants 14.5 instead of 29/2


You're on your own now, buddy.

Answer 15

could it be you made a mistake?

Answer 16

I think the question is poorly worded and it is hard to say. Obviously the displacement is negative (you end up to the left), but the net distance traveled might mean the absolute value of the displacement. Whoever wrote the question does not do physics.

Answer 17

I think it is asking for the absolute value

Answer 18

I am not so sure about that, displacement, unlike distance, includes direction.