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].
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
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
I am still getting it wrong
since they want displacement, ignore sign changes.
∫[0,5] -t^2+3t-2 dt = -86/5
-86/5 isn't right
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.
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
its asking for total distance traveled so the answer wont be negative
Yes, there must be sign changes
It asked for the NET distance, not the TOTAL distance
Yes total is to the left