Find the distance, in feet, a particle travels in its first 4 seconds of travel, if it moves according to the velocity equation v(t)= -t2 + 4 (in feet/sec).

Question

a) 55/3

b) 16/3

c) 16

d) 12

Answer 1

v(t) = -t^2 + 4

0<=t < 2, v(t) > 0, particel travels in positive direction

at t = 2, v(t) = 0

t > 2, v(t) < 0, particel travels in negative direction

distance = distance traveled in positive direction + negative direction

r1 = ∫₀² v(t) dt = 16/3

r2 = ∫₂⁴ v(t) dt = -32/3

d = |r1| + |r2| = 48/3 = 16