Calculus

The image shows a velocity-time graph where: - The y-axis (velocity) ranges from -6 to 5 - The x-axis (time) ranges from 0 to 11 - Initial velocity is known - Initial position is known From the graph, the velocity function appears to be: - v = 5 from t = 0 to t = 1 - v decreases linearly from 5 to -6 from t = 1 to t = 5 - v = -6 from t = 5 to t = 9 - v increases from -6 to 0 from t = 9 to t = 11 Given a velocity-time graph with initial velocity and initial position known, find the position function x(t). Find the displacement starting from 5 seconds to 11 seconds

Step-by-step solution with explanation

Final Answer

The displacement from t = 5 to t = 11 is units.

Step-by-step solution

1

Identify velocity segments from 5 to 11

We only care about t = 5 to t = 11. From t = 5 to t = 9, velocity is constant at -6. From t = 9 to t = 11, velocity rises linearly from -6 to 0, giving a slope of (0 - (-6))/(11 - 9) = 3.
2

Displacement equals area under velocity curve

Displacement from t = 5 to t = 11 is the total area between the velocity curve and the time axis. We split it at t = 9 where the shape changes.
3

Calculate area of constant segment (5 to 9)

This is a rectangle below the time axis. Width is 4, height is -6, so area is -24. Negative area means the object moved in the negative direction.
4

Calculate area of triangle segment (9 to 11)

From t = 9 to t = 11, the velocity goes from -6 to 0. This forms a triangle below the axis. Base = 2, height = 6 (below axis, so negative), giving area = -6.
5

Add both areas for total displacement

We add the two pieces together. Both are negative, meaning the object moved 30 units in the negative direction between t = 5 and t = 11.

Understanding this problem

Learning Insight

Displacement is always the signed area under a velocity-time graph. Regions below the t-axis count as negative displacement because the object moves backward (or in the negative direction). This is a direct application of the relationship x(t) = ∫v(t) dt.

Quick Tip

Break the velocity graph into simple shapes — rectangles and triangles. Area = base × height for rectangles, and ½ × base × height for triangles. Just remember to make the area negative if the shape is below the time axis.

Common Mistake

Students often use the total distance formula (adding absolute values) instead of signed area, which gives +30 instead of -30. Displacement requires signed area; total distance requires unsigned area.