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Distance and Displacement

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Distance and Displacement 
The area underneath the velocity curve of an object represents its distance traveled. If at some point the object moves backwards, then its velocity will be negative (subtracting some distance). In such cases we are measuring displacement - the distance between the initial and final position. 


Displacement
If v(t)v(t)v(t)is the velocity of a particle on the interval [a,b][a,b][a,b]then the displacement of the particle is given by
∫abv(t) dt= displacement\boxed{\int_a^b v(t)\, dt=\text{ displacement}}∫ab​v(t)dt= displacement​

Total Distance 
If v(t)v(t)v(t)is the velocity of a particle on the interval [a,b][a,b][a,b]then the total distance of the particle is given by
∫ab∣v(t)∣ dt= total distance traveled\boxed{\int_a^b|v(t)|\, dt=\text{ total distance traveled}}∫ab​∣v(t)∣dt= total distance traveled​


Watch Out!
Displacement is also called net distance; don't mix up net distance and total distance.

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Example: Distance and Displacement
A particle moves along a line so that its velocity at time t is v(t)=t2−t−6v(t)=t^2-t-6v(t)=t2−t−6, in meters per second. 
Find the displacement of the particle during the time period 1≤t≤41 \leq t \leq 41≤t≤4, and then find the total distance traveled during this time period.

Displacement is net distance, which is just the integral of the velocity function over the time interval:
displacement =∫14v(t) dt=∫14(t2−t−6) dt=[t33−t22−6t]14=−92\text{displacement }=\int_1^4v(t)\, dt = \int_1^4(t^2-t-6)\, dt=\left[\frac{t^3}{3}-\frac{t^2}{2}-6t \right]_1^4=-\frac{9}{2}displacement =∫14​v(t)dt=∫14​(t2−t−6)dt=[3t3​−2t2​−6t]14​=−29​
The negative sign means that the particle's position at time t=4t=4t=4 was 9/2 or 4.5 meters to the left of its position at time t=1t=1t=1.

To find total distance, we need the absolute value of the velocity function. We need to find where the velocity function equals zero, and where it's negative (below the x-axis), we will make it positive:
t2−t−6=(t−3)(t+2)t^2-t-6=(t-3)(t+2)t2−t−6=(t−3)(t+2).

At time t=1t=1t=1, this is negative; and it becomes positive at time t=3t=3t=3, and then stays positive for the rest of the time interval. So we need to split the integral at t=3t=3t=3, and flip the sign for the interval 1≤t≤31 \leq t \leq 31≤t≤3:
∫14∣v(t)∣ dt=∫13∣v(t)∣ dt+∫34∣v(t)∣ dt=∫13−v(t) dt+∫34v(t) dt=∫13−(t2−t−6) dt+∫34(t2−t−6) dt=∫13(−t2+t+6) dt+∫34(t2−t−6) dt=[−t33+t22+6t]13+[t33−t22−6t]34=616 meters \begin{array}{rl}\displaystyle
\int_1^4 |v(t)|\, dt & = \int_1^3|v(t)|\, dt + \int_3^4 |v(t)|\, dt\\[+1em]
\displaystyle&=\int_1^3-v(t)\, dt + \int_3^4 v(t)\, dt \\[+1em]
&=\int_1^ 3-(t^2-t-6)\, dt + \int_3^4(t^2-t-6)\, dt\\[+1em]
&=\int_1^3 (-t^2+t+6)\, dt + \int_3^4(t^2-t-6)\, dt\\[+1em]
&=\left[ -\frac{t^3}{3}+\frac{t^2}{2}+6t \right]_1^3+\left[ \frac{t^3}{3}-\frac{t^2}{2}-6t \right]_3^4\\[+1em]
&=\frac{61}{6} \text{ meters}
\end{array}∫14​∣v(t)∣dt​=∫13​∣v(t)∣dt+∫34​∣v(t)∣dt=∫13​−v(t)dt+∫34​v(t)dt=∫13​−(t2−t−6)dt+∫34​(t2−t−6)dt=∫13​(−t2+t+6)dt+∫34​(t2−t−6)dt=[−3t3​+2t2​+6t]13​+[3t3​−2t2​−6t]34​=661​ meters​

Practice: Distance and Displacement 
The velocity function, in meters per second, of a particle moving alone a line is given by 
v(t)=t2−2t−8v(t)=t^2-2t-8v(t)=t2−2t−8
For the time interval 1≤t≤61 \leq t \leq 61≤t≤6, find 
(a) the displacement of the particle
(b) the distance traveled by the particle

(a) (don't include units)

(b) (don't include units)

I don't know

Practice: Displacement 
The following graph represents the velocity of a particle in m/s, find the displacement in meters of the particle from t=0 to t=5.

(Don't Include Units)

I don't know