The velocity function (in meters per second) is given for a particle moving along a line.

Question

v(t) = 3t − 7, 0 ≤ t ≤ 3
(a) Find the displacement.
-7.5 m
(b) Find the distance traveled by the particle during the given time interval.

Answer 1

negative v at start, positive at 3 so reverses direction during the 3 seconds at t = 7/3
v = 3 t - 7
for vector displacement
x = (3/2)t^2 - 7 t + c
at t = 3, x = 27/2 - 21 + c
at t = 0, x = c
so
displacement = -15/2 = -7.5 agree
then
for distance
integrate from t = 0 to t = 7/3
x = (3/2)t^2 - 7 t + c = (3/2)(49/9) - 49/3 = -49 /6 , negative motion
then integrate from t = 7/3 to t = 3
at t= 3
x = 27/2 - 21 + c
at t = 7/3
x = (3/2)(49/9) -49/3+ c = -49/6 +c
motion = 27/2 -21+49/6 = + 2/3 positive motion
total distance moved = 49/6 + 4/6 = 53/6