calculus homework

Question

A particle moves along the x-axis search that it’s position at any time (t) is given by the equation x(t)= t^2— 2t +9. Find the distance the particle moves from t= -3 to t=4

Answer 1

To solve this problem, we must find the intervals on which the particle is moving to the left and to the right. Since t^2 - 2t + 9 = (t - 1)^2 + 8, the particle is moving to the left for t < 1 and to the right for t > 1. Therefore, the total distance the particle moves is (x(-3) - x(1)) + (x(4) - x(1)) = 24 - 8 + 17 - 8 = 25.

We could also find the intervals on which the particle is moving left and right by finding the derivative of x(t), which is 2t - 2. This is clearly negative for t < 1 and positive for t > 1, but my method above uses no calculus.

Answer 2

if the question is asking for displacement, the distance from starting to end point, then

d = x(-3) - x(4) = 9+6+9 - (16-8+9)= 24-17 = 7

d = 7

but if it's asking for the total distance of movement along the x axis, back and forth

check first where velocity = 0, where the direction of movement changes

v(t) = x'(t) = 2t-2= 0,

t = 2/2,

t= 1

then the total distance traveled

d = x(-3) - x(1) + x(4) -x(1) = 24-8 + 17-8 = 16+9 = 25

d = 25