What is the fundamental difference between the calculation of distance and displacement using integrals?

Displacement is the integral of velocity, $\int v(t) dt$, which allows positive and negative motion to cancel out. Distance is the integral of speed, $\int |v(t)| dt$, which treats all motion as positive accumulation.

Under what specific condition will the total distance traveled be exactly equal to the magnitude of the displacement?

This occurs only when the object moves in a single direction for the entire duration of the interval. Mathematically, this means the velocity $v(t)$ does not change sign (it remains either $\ge 0$ or $\le 0$).

How does the concept of speed relate to the concept of velocity in a calculus context?

Speed is the magnitude of the velocity vector, expressed as $|v(t)|$. While velocity indicates both how fast and in what direction an object moves, speed only indicates the rate of motion.

What is a common error when using the absolute value in the distance formula?

A common mistake is placing the absolute value outside the integral, as in $|\int v(t) dt|$. This calculates the magnitude of the net displacement rather than the total distance traveled.

Why is it necessary to find the roots of $v(t)$ when calculating distance by hand?

The roots indicate where the object stops or changes direction. To calculate distance, you must split the integral at these points to ensure that negative velocity segments are manually converted to positive values.

If an object has a negative velocity, what happens to the total distance traveled?

The total distance continues to increase. Because distance is the integral of speed (the absolute value of velocity), even negative velocity contributes a positive amount to the total path length.

Define 'Speed' in terms of calculus operations.

Speed is the absolute value of the velocity function, $|v(t)|$, or the first derivative of the distance function with respect to time.

State the definite integral formula for total distance traveled between time $a$ and $b$.

The formula is $\text{Distance} = \int_{a}^{b} |v(t)| \, dt$. This represents the area under the speed-time graph.

What does the area under a speed-time graph represent?

The area under a speed-time graph represents the total distance traveled. This is distinct from the area under a velocity-time graph, which represents displacement.

If $v(t) = t - 3$, why is the distance from $t=0$ to $t=4$ not simply the integral of $v(t)$?

Because $v(t)$ is negative from $t=0$ to $t=3$ and positive from $t=3$ to $t=4$. Integrating $v(t)$ directly would subtract the first area from the second, giving displacement instead of distance.

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Unit 8: Applications of Integration

Distance & Speed

Summary

In calculus, distance and speed are the scalar magnitudes of the vector quantities displacement and velocity. While displacement measures the net change in position, distance represents the total path length traveled, calculated by integrating the speed (the absolute value of velocity) over a specific time interval.

1. Definition & Core Concepts

Speed is defined as the magnitude of the velocity vector, denoted as $|v(t)|$ . It is a scalar quantity that represents how fast an object is moving regardless of its direction.
Distance is the total path length covered by an object during its motion, represented as the magnitude of displacement $|s|$ . Unlike displacement, distance can never decrease over time.
Scalar vs. Vector: Velocity and displacement are vectors because they include direction (positive or negative), whereas speed and distance are scalars and are always non-negative.

A graph showing a velocity curve crossing the x-axis, illustrating that distance is the sum of the absolute areas under the curve.

2. Underlying Principles

3. Methods & Techniques

4. Key Distinctions

5. Exam Strategy & Tips

Check for Sign Changes: Always check if the velocity function crosses the x-axis within the given bounds. If it does, a simple integral of velocity will yield displacement, not distance.
Units Consistency: Ensure that the units of speed (e.g., meters/second) and time (e.g., seconds) match the required units for distance (e.g., meters).
Sanity Check: Total distance must always be greater than or equal to the magnitude of displacement. If your distance is smaller than your displacement, an error has occurred in the calculation.
Absolute Value Placement: In the distance formula, the absolute value signs are inside the integral. Placing them outside, $|\int v(t) dt|$ , calculates the magnitude of displacement, which is incorrect for total distance.