Let $x(t)$ be the position of an object moving along a number line. Suppose that the velocity of the object is $\frac{dx}{dt}$ = $7 + 5$cos$($t/4$)$

Question

Let $x(t)$ be the position of an object moving along a number line. Suppose that the velocity of the object is $\frac{dx}{dt} = 7 + 5\cos(t/4)$

What is the displacement of the object between $0\leq t\leq 2\pi$? What is the distance travelled between $0\leq t\leq 2\pi$?

I first found the function of the position (by taking the anti-derivative since we aren't allowed to integrate yet in class).

I got to $x(t) = 7t + 20\sin(t/4)$ + $c$, and then I did $x(2\pi) - x(0) = 14\pi + 20$ as the displacement.

I am unsure how to get the distance. Since the object is moving on a number line, wouldn't the distance and displacement be the same?

(Please let me know if my displacement is wrong )

Answer 1

Mind that the object is not moving straight. If you go with your car outside and you drive 100m in north direction, then go back in south direction 50m, then turn again and go north for 100m the total distance you have done is 250m but you are 200m away from your starting point. So displacement can be different of travel distance. However in that case, as the derivative is positive you don't change direction so the distance travelled is the displacement.

Answer 2

In general, the movement can be back and forth, so displacement is not the same as the distance traveled. However, in this case $x'(t)=7+5\sin (t/4)$ is always positive which means the movement is only in the positive direction Hence, distance is same as displacement.

Answer 3

The object moving on a number line does not necessarily mean that the displacement equals the distance travelled, since it might have gone through the same section multiple times. However, in this case, the velocity is always positive (always $\cos{(t/4)} > -\frac{7}{5}$), so this does not happen. Therefore, the distance travelled does equal to the displacement.