The graph shows the velocity $v(t)$ of an object moving along a straight line as a function of time $t$, measured in hours. The areas under the velocity curve and the $t$-axis represent displacement (which can be negative) or total distance traveled (always positive). The given areas represent the integral of the velocity function over different time intervals.

Information from the graph:

• From $t = 0$ to $t = 1$, the area under the curve is 12 (positive area).
• From $t = 1$ to $t = 3$, the area is 16 (negative area, as the velocity is below the $t$-axis).
• From $t = 3$ to $t = 5$, the area is 10 (positive area).

Displacement Calculation:

Displacement is the net change in position, which is the algebraic sum of the areas. Positive areas correspond to forward motion, and negative areas correspond to backward motion.

$\text{Displacement} = 12 - 16 + 10 = 6 \text{ miles}.$

Distance Traveled Calculation:

The total distance traveled is the sum of the absolute values of the areas, since distance is always positive.

$\text{Total Distance} = |12| + |16| + |10| = 12 + 16 + 10 = 38 \text{ miles}.$

Thus:

• Displacement: $6$ miles.
• Total Distance Traveled: $38$ miles.

Would you like more details on how these integrals are interpreted, or have any questions? Here are some questions to consider:

1. What is the difference between displacement and total distance traveled?
2. How does the sign of the velocity affect the displacement and distance?
3. Can displacement ever exceed the total distance traveled?
4. How are areas under the curve interpreted when velocity is negative?
5. How would the calculation change if the object changed direction multiple times?

Tip: The area under the velocity curve between two points represents the displacement over that time interval.