A particle moves along the x-axis so that its position at any time t ≥ 0 is given by x(t) = −t^3 +7t^2 −14t +8.

Question

A particle moves along the x-axis so that its position at any time t ≥ 0 is given by x(t) = −t^3 +7t^2 −14t +8.

Find the total distance traveled during the first two seconds.

Answer 1

If the question was asking for the displacement, it would be a simpler problem because we would be able to do x(2) - x(0) and get our answer.

However, distance is always positive, and therefore distance is the absolute value of displacement. Therefore, we need to first find out if the particle switches directions at all during the interval [0,2]. The particle would switch directions if the velocity switch from + to - or from - to + during that time interval.

Velocity is the derivative of position, so we will find the velocity function, and set it equal to zero and solve for t. We set it equal to zero because it can only switch from + to - or from - to + by going through v(t) = 0.

x'(t) = v(t) = -3t^2 + 14t - 14 = 0

Using the quadratic formula, we get only one zero between t = 0 and t = 2, and that zero (critical number) is @ t = 1.4514162 seconds.

Furthermore, to the left of that critical number, v(1) is negative.

And to the right of that critical number, v(2) is positive.

Because the velocity switches signs, the position does switch directions at t = 1.4514162 seconds.

Therefore, the total distance traveled = (total distance from t = 0 to t = 1.4515) + (total distance traveled from t = 1.4515 to t = 2).

Total distance traveled = -[x(1.4515) - x(0)] + [x(2) - x(1.4515)]....notice that I multiplied the first brackets by negative one to ensure that that quantity would become positive (otherwise it would've been negative, but distance must be positive).

Total distance traveled = - [-.6311303 - 8] + [0 - (-.6311303)]

Total distance traveled = 9.26226 units.