Particle movement: total distance

Question

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?

a) 168

b) 171

c) 175

d) 375

Answer 1

The total distance travelled is different than the change in position from t = 0 to t = 7.

If you find the first derivative of x(t) and set it equal to zero you will find that the velocity of the particle is 0 when t = 5, and that it is travelling in the negative direction from t = 0 to t = 5 and in the positive direction from t = 5 to t = 7.

So, you have to find |x(5) - x(0)| to get the distance travelled in the negative direction.

Then determine |x(7) - x(5)| to get the distance travelled in the positive direction.

Add those two values together to get total distance travelled.

This is the same as:∫₀⁷ | v(x)|dx, where you have to break this into two separate integrals to evaluate manually.

Visit:

desmos.com/calculator/fpwmqozifu

to confirm.

Answer 2

Hi Bob M.,

The distance traveled by the particle will be the absolute value of the final position of x (x_f) minus the initial position of x (x_i) or d = |x_f - x_i|.

Or you can use the distance formula, d = √[(x₂-x₁)²+(y₂-y₁)²], with y₂=0 and y₁=0, same difference.

With x(t)= 2t³- 9t²- 60t + 4 and 0 ≤ t ≤ 7:

x_f(7) = 2(7)³ - 9(7)² - 60(7) + 4 = -171, (the particle moves in the negative x-direction).

x_i(0) = 4, (the starting point on the x-axis)

d = |-171 - 4| = 175

I hope this helps, Joe.