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### Course: Calculus 2 > Unit 4

Lesson 2: Straight-line motion- Motion problems with integrals: displacement vs. distance
- Analyzing motion problems: position
- Analyzing motion problems: total distance traveled
- Motion problems (with definite integrals)
- Analyzing motion problems (integral calculus)
- Worked example: motion problems (with definite integrals)
- Motion problems (with integrals)
- Average acceleration over interval

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# Motion problems (with definite integrals)

Definite integrals are commonly used to solve motion problems, for example, by reasoning about a moving object's position given information about its velocity. Learn how this is done and about the crucial difference of velocity and speed.

Motion problems are very common throughout calculus. In differential calculus, we reasoned about a moving object's velocity given its position function. In integral calculus we go in the opposite direction: given the velocity function of a moving object, we reason about its position or about the change in its position.

## Thinking about velocity, speed, and definite integrals

Say a particle moves in a straight line with velocity $v(t)=5-t$ meters per second, where $t$ is time in seconds.

When the velocity is positive it means the particle is moving forward along the line, and when the velocity is negative it means the particle is moving backwards.

Say we are asked for the particle's $t=0$ seconds and $t=10$ seconds. Since the velocity is the

*displacement*(i.e. the change in its position) between*rate of change*of the particle's position, any change in the position of the particle is given by a definite integral.Specifically, we are looking for ${\int}_{0}^{10}v(t){\textstyle \phantom{\rule{0.167em}{0ex}}}dt$ .

Interestingly, the displacement is ${\int}_{0}^{10}v(t){\textstyle \phantom{\rule{0.167em}{0ex}}}dt=0$ meters. (You can see how the two areas in the graph are equal in size and opposite in sign).

The displacement being $0$ means that the particle was at the same position at times $t=0$ and $t=10$ seconds. This makes sense when you see how the particle first moves forwards and then backwards, so it gets back to where it started.

Nevertheless, the particle

*did move*. Say we want to know the*total distance*the particle covered, even though it ended up in the same place. Can definite integrals help us with that?Yes, they can. To do that, we will use a clever manipulation. Instead of looking at the particle's $v$ , we will look at its $|v|$ (i.e. the absolute value of $v$ ).

*velocity**speed*Speed describes how fast we go, while velocity describes how fast

*and*in which direction. When the movement is along a line, velocity might be negative, but speed is always positive (or zero). So speed is the absolute value of velocity.Now that we know the particle's ${\int}_{0}^{10}|v(t)|{\textstyle \phantom{\rule{0.167em}{0ex}}}dt$ .

*speed*at any moment, we can find the*total distance*it covered by calculating the definite integralThis time the result is the positive value $25$ meters.

#### Remember: velocity vs. speed

Velocity is rate of change in position, so its definite integral will give us the

*displacement*of the moving object.Speed is the rate of change in

*total distance*, so its definite integral will give us the total distance covered, regardless of position.## Finding the actual position using definite integrals and initial conditions

Some motion problems ask us to find the actual position of the particle at a certain point in time. Remember that a definite integral can only give us the

*change*in the particle's position. In order to find an actual position of the particle, we will need to use*initial conditions*.*Want more practice? Try this exercise.*

## Summary: Three possible prompts in motion problems involving definite integrals

Motion problems require definite integrals when we're given the moving object's

*velocity*and are asked about its*position*. There are three types of possible problems:Type | Common prompt | Appropriate expression |
---|---|---|

Displacement | "What is the particle's displacement between... and..." or "What is the change in the particle's position between... and..." | |

Total distance | "What is the total distance the particle has traveled between... and..." | |

Actual position | "What is the particle's position at..." |

## Want to join the conversation?

- How do I know at what time a particle returns to the origin?(9 votes)
- When the area under the curve above the x- axis is equal to the area under the curve below the x-axis on the given interval for a velocity-time graph.

Note- This actually gives us the condition for zero displacement. So when the starting point of the particle is the origin, then zero displacement implies that the final point is also the origin. In other words, the particle returns to the origin.(19 votes)

- In Problem #1, it says that the particle moves in a straight line. But the position function includes x^3, which is not a straight-line function. Is the statement that the particle moves in a straight line incorrect?(7 votes)
- The particle moves along a line, but the particle's position on the line changes over time in a non-linear fashion.(16 votes)

- How do I figure out the greatest distance between the particle and the origin?(6 votes)
- It depends a lot on the path taken by the particle.

For example, if it (the particle) performs oscillatory motion about the origin, the greatest displacement would be when it is at an extreme position (when its velocity is zero). However, if it moves in a circular path with the origin as the center, the greatest distance ( also the distance at all times) would be the length of the radius. So if you could be a bit more specific about what kind of path it is, I think I may be able to help you better.(1 vote)

- why it is so confusing ?(1 vote)
- Try to cover "One / Two dimensional motion" from both "Physics" and "Physics AP" playlists (they complement each other) and this will become trivial to you.

They are not calculus based, so you'll also have to know how position (x), velocity (v), acceleration (a), and jerk (j) can be linked together using derivatives and integrals.

(derivative of one in respect to time gives the next one)

position (x) -> velocity (v) -> acceleration (a) -> jerk (j).

d/dt(x) = v (i.e. derivative of position is velocity)

d/dt(d/dt(x)) = d/dt(v) = a

d/dt(d/dt(d/dt(x))) = d/dt(a) = j

(integral of one in the chain gives the previous one)

position (x) <- velocity (v) <- acceleration (a) <- jerk (j).

Integration works the opposite way, letting you get, say, position (x) from velocity (v), i.e. integral of velocity is position.

Update: it's turned out this is covered in the next video.(9 votes)

- Why in acceleration problems do we not us + C in order to find the entire integral?(2 votes)
- These are definite integrals, so we know the exact answer(7 votes)