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).

Step-by-step solution with explanation

Final Answer

The object ends at , meaning it is 27 units behind its starting position.

Step-by-step solution

1

Break velocity into four interval pieces

We read each segment from the graph. A straight line from to has slope . A straight line from to has slope . Constant pieces have slope 0.
2

Integrate segment 1: t = 0 to 1

Position equals initial position plus the integral of velocity. On this segment velocity is the constant 5, so integration is straightforward. At : .
3

Integrate segment 2: t = 1 to 5

We start from the position at and integrate the linear velocity expression. Let to keep the algebra clean. At : .
4

Integrate segment 3: t = 5 to 9

Velocity is the constant here, so position decreases linearly. At : .
5

Integrate segment 4: t = 9 to 11

Velocity rises linearly from back to . We integrate, starting from . At : .
6

Write the complete piecewise position function

Each piece picks up exactly where the previous piece ended, so is continuous everywhere. Plug in any known value of (initial position) to get a fully numerical answer. The function is unique — there is only one solution.

Understanding this problem

Learning Insight

Position is the integral of velocity because velocity tells you how fast position is changing. Each area region under the velocity-time graph — positive or negative — adds to or subtracts from the position. This is why a negative velocity region (like to ) moves the object backward.

Quick Tip

On a velocity-time graph, the NET displacement over any interval is just the signed area under the curve. Rectangles and triangles are your best friends here — split each linear segment into those two shapes to find area fast without full integration.

Common Mistake

Students often forget to carry over the ending position of one segment as the starting position of the next. Each piece must begin where the previous piece left off — using as the start for every segment gives a completely wrong function after the first interval.