Total distance traveled by a particle

Question

A particle moves according to the equation of motion, $$s(t)=t^2-2t+3$$ where $s(t)$ is measured in feet and t is measured in seconds. Find the total traveled distance in the first $3$ seconds.

Answer 1

The key idea here is that you need to know if and when the particle changes direction of motion. Therefore differentiate the displacement to obtain the velocity and set this to zero.

It is readily seen that the velocity is zero when $t=1$.

The initial position at $t=0$ is $s=3$. At $t=1, s=2$. Distance travelled so far is $1$.

At $t=3, s=6$, so further distance travelled is $6-2=4$.

Therefore total distance is $5$.

Answer 2

The total traveled distance between $t=0$ and $t=3$ is the length of the image of $s_{|_{[0,3]}}$, which is $$ L=\int_0^3|s'(t)|dt=\int_0^3|2t-2|dt=\int_0^12-2tdt+\int_1^32t-2dt=5. $$

Edit: If you don't know the arc length formula another approach to calculate the traveled distance of a particle satisfying a law of motion $s(t)$ between $t=a$ and $t=b$ would be to calculate the critical points in $(a,b)$, say $t_1<\cdots<t_k$, and then calculate the traveled distance as $|s(t_1)-s(a)|+|s(t_2)-s(t_0)|+\dots+|s(b)-s(t_k)|$. For your specific example there is only the critical point $t=1$, so $L=|s(1)-s(0)|+|s(3)-s(1)|=1+4.$

Answer 3

Another method (avoiding the use of derivatives and integrals) is the following: Let's plot the graph for $s(t)$:

From the above we can see that the particle changes direction at $t=1$. This is the point where the minimum value of the above parabola (i.e. of the diplacement from the origin) appears: $s(1)=2$. So, for the total distance: $$ |s(1)-s(0)|+|s(3)-s(1)|=|2-3|+|6-2|=1+4=5 $$