A particle moves along the x-axis so that at time t ≥ 0 its position is given by x(t)=2t^3−9t^2−60t+4. What is the total distance traveled by the particle over the time interval 0≤t≤7 ?

Answer 1

To find the total distance traveled by this particle you want to consider that this particle could move in only 2 directions: in the positive x direction (to the right) or in the negative x direction (to the left). What you need to figure out is at which times it turns around and changes direction as well as its position at those times. Then you can calculate the distance between those points and add it all up.

To find the times when the particle changes direction, you just need to find the critical numbers of the function x(t). These would be the possible times when the particle changes direction.

x(t) = 2t³ - 9t² - 60t + 4

x'(t) = 6t² - 18t - 60 = 6 (t² - 3t - 10) = 6 (t -5) (t + 2)

0 = 6 (t -5) (t + 2)

t = 5, -2 (Since 0 ≤ t ≤ 7, we only need t = 5)

Then you can find the position of the particle at these times. We will also need to find its position at our end points: t = 0, 7. Basically all we are doing here is finding the global max/min values of the function up to this point.

x(0) = 4

x(5) = -271

x(7) = -171

So from t=0 to t=5, the particle moved 275 units in the negative direction. Then from t=5 to t=7, the particle moved 100 units in the positive direction. Therefore, between t=0 to t=7, the particle moved a total of 375 units.

Answer 2

Let find velocity: v(t) = x'(t) = 6t² - 18t - 60;

Total distance s = ∫₀⁷abs(6t² - 18t - 60)dt = 6∫₀⁷abs(t² - 3t -10)dt = 6[∫₀⁵(- t² + 3t + 10)dt + ∫₅⁷(t² -3t - 10)dt] =

6[(- t³/3 + 3t²/2 + 10t)₀⁵ + (t³/3 - 3t²/2 -10t)₅⁷] = 6[(- 125/3 + 75/2 + 50) + (343/3 - 147/2 - 70) - (125/3 - 75/2 - 50)] = - 250 + 225 + 300 + 686 - 441 - 420 - 250 + 225 + 300 = 375