Question

Multiple Choice A truck moves with positive velocity \(v(t)\) from time \(t=3\) to time \(t=15 .\) The area under the graph of \(y=v(t)\) between 3 and 15 gives (A) the velocity of the truck at \(t=15\) . (B) the acceleration of the truck at \(t=15\) . (C) the position of the truck at \(t=15\) . (D) the distance traveled by the truck from \(t=3\) to \(t=15\) (E) the average position of the truck in the interval from \(t=3\) to \(t=15 .\)

Short answer

The correct answer choice is (D) The distance traveled by the truck from \( t = 3 \) to \( t = 15 \). The area under the graph of velocity-time between two points gives the distance travelled between those points in time, if the velocity is always positive.

Step-by-step solution

Rule out the incorrect answers

Let's analyze each option individually to understand which ones can be ruled out. (A) The velocity of the truck at \( t = 15 \) is not correct because the area under the velocity-time graph does not give instant velocity, but total distance covered. (B) Similarly, the acceleration of the truck at \( t = 15 \) is ruled out as well because acceleration is the rate of change of velocity and not represented by the area under the velocity-time graph. (C) The position of the truck at \( t = 15 \) is also not correct because without further information about the initial position, the area under the graph won't give the final position.

Analyze the remaining options

Now, let's analyze the remaining options. (D) The distance traveled by the truck from \( t = 3 \) to \( t = 15 \) seems promising because the area under the velocity vs. time graph indeed gives the displacement (distance travelled, in this case, because of the positive velocity) over the time interval. (E) The average position of the truck in the interval from \( t = 3 \) to \( t = 15 \) is incorrect because the average position would require knowledge of the initial and final positions, which we don't have.

Choose the correct answer

From the detailed analysis in Steps 1 and 2, it's clear that the correct answer is (D) The distance traveled by the truck from \( t = 3 \) to \( t = 15 \). This problem demonstrates important concepts in physics and calculus: the area under a velocity-time graph represents the distance travelled over the given time interval, if the velocity is always positive.

Key concepts

Distance Traveled Calculus

When it comes to understanding movement, 'distance traveled' is a key concept heavily relying on calculus. If you're faced with a velocity-time graph and need to know how far an object has traveled, calculus comes to the rescue.

Consider the graph of velocity over time. Calculus teaches us that to find the distance traveled, you'll want to look at the area under the curve of this graph. This is because the integral of the velocity function with respect to time gives you the object's displacement. For a positive velocity, this displacement is equivalent to the distance traveled.

It's like adding up tiny slices of distance the object travels in each minuscule segment of time—these add up to the total distance. So, when given a problem where you need to determine how much ground something has covered over time, remember: integrate the velocity over the given interval.

Velocity-Time Relationship

The connection between velocity and time is fundamental in physics. A velocity-time graph showcases how an object's speed and direction of travel change over time. In calculus, this graph can provide a wealth of information.

For example, the slope of the velocity-time graph gives you the object's acceleration—a steeper slope indicates a greater change in velocity or, in other words, stronger acceleration. Now, if you maintain a positive velocity throughout, as in our truck example, your graph stays above the time-axis.

What's even more intriguing is the relationship between the shape of the curve and the object's motion. A straight line means the object travels at a constant velocity, while a curve signals changing speeds. Through calculus, the velocity-time graph is not just a picture but a storyteller of the object's journey over time.

Calculus in Physics

Calculus is like a secret decoder ring for understanding the language of physics. It's a tool to analyze and understand the dynamics of objects, from celestial bodies in space to cars racing on a highway.

Through differentiation, you can find the rate of change of quantities, grasping concepts like velocity (change in position over time) and acceleration (change in velocity over time). Meanwhile, integration, which is essentially summation over continuous quantities, lets you calculate things like distance traveled from velocity or the area under a curve.

This branch of mathematics is indispensable in physics because it allows precise predictions and insights into motion, forces, and energy. It is the foundation for understanding how the universe operates on every scale.

Positive Velocity

In the context of our truck scenario, 'positive velocity' is like having a constant forward motion. As long as the velocity remains positive, the truck is moving in one direction—no backing up or standing still.

A positive velocity simplifies our calculations for travel distance, as we don't need to account for any changes in direction. It's a straight shot from point A to B, so to speak. Thus, on a velocity-time graph, positive velocity translates into the graph staying above the time (horizontal) axis, and the total distance traveled is just the area under the curve between your start and end times.

When teaching the concept of positive velocity, it's helpful to stress that while speed tells us how fast an object is moving, velocity also gives us the direction - and in these problems, that direction remains constant and forward.