A particle moves along a line with velocity v=3t^2-6t. The total distance traveled from t= 0 to t= 3 equals

Question

A particle moves along a line with velocity v=3t²-6t. The total distance traveled from t= 0 to t= 3 equals

(A) 9

(B) 4

(C) 2

(D) 16

(E) none of these

What is the net change of the particle?

A. 2

B. 4

C. 9

D. 16

E. None

Answer 1

v(t) = 3t^2 - 6t

distance = the integral of v(t) = s(t)

s(t) = t^3 -3t^2

s(0) = 0 so there is no constant term in s(t)

s(1) =1-3 = -2

s(2) =8-12 = -4

s(3) = 3^3 -3(3)^2 = 0

3t^2-6t = 0, t=2 is when it changes direction, when v=0

the particle went backwards 4 in 2 seconds then forward 4 in the next second

The answer is B if 4 is the distance traveled, or 8 if you add 4 in one direction to 4 in the other, or 0

if you add -4 to +4 as the distance. Answer is either B: 4, or E none of the above

Net change is zero or 4-4=0

E None