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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, left parenthesis, t, right parenthesis, equals, 5, minus, t meters per second, where t is time in seconds.
Function v is graphed. The horizontal axis is represented by t. t goes from 0 to 10. The graph is a line. The line starts at (0, 5), intersects the t axis at (5, 0), and ends in the fourth quadrant.
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 displacement (i.e. the change in its position) between t, equals, 0 seconds and t, equals, 10 seconds. Since the velocity is the 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 integral, start subscript, 0, end subscript, start superscript, 10, end superscript, v, left parenthesis, t, right parenthesis, d, t.
The previous graph of function v is displayed. The area between the graph and the t axis is shaded from t = 0 to t = 10. This consists of two separate areas, to the left and to the right of t = 5.
Interestingly, the displacement is integral, start subscript, 0, end subscript, start superscript, 10, end superscript, v, left parenthesis, t, right parenthesis, d, t, equals, 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, equals, 0 and t, equals, 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.
A dot representing a particle is plotted at the left end of a horizontal line, where t = 0 and v = 5. The particle moves to a maximum point to the right, then back left to the start. As it moves right, t increases to 5 and v decreases to 0. As it moves back left, t increases to 10 and v decreases to negative 5.
This is a simulation of the particle's movement from t, equals, 0 seconds to t, equals, 10 seconds (created with Geogebra).
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 velocity v, we will look at its speed vertical bar, v, vertical bar (i.e. the absolute value of v).
The previous graph of function v is displayed. A second function, the absolute value of v is graphed. The graph consists of two line segments. The first moves downward from (0, 5) to (5, 0). The second moves upward from (5, 0) through (10, 5).
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 speed at any moment, we can find the total distance it covered by calculating the definite integral integral, start subscript, 0, end subscript, start superscript, 10, end superscript, vertical bar, v, left parenthesis, t, right parenthesis, vertical bar, d, t.
The previous graph of function absolute value of v is displayed. The area between the graph and the t axis is shaded from t = 0 to t = 10. This consists of two separate areas, to the left and right of t = 5.
This 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.
Problem 1
A particle moves in a straight line with velocity v, left parenthesis, t, right parenthesis, equals, minus, t, squared, plus, 8 meters per second, where t is time in seconds. At t, equals, 2, the particle's distance from the starting point was 5 meters. What is the total distance the particle has traveled between t, equals, 2 and t, equals, 6 seconds?
Which expression should Alexey use to solve the problem?

Problem 2
A particle moves in a straight line with velocity v, left parenthesis, t, right parenthesis meters per second (graphed), where t is time in seconds. At t, equals, 1, the particle's distance from the starting point was 2 meters in the positive direction. What is the particle's displacement between t, equals, 1 and t, equals, 6 seconds?
Which expression should Madelyn use to solve the problem?

## 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.
Problem 3
A particle moves in a straight line with velocity v, left parenthesis, t, right parenthesis, equals, square root of, 3, t, minus, 1, end square root meters per second, where t is time in seconds. At t, equals, 2, the particle's distance from the starting point was 8 meters in the positive direction. What is the particle's position at t, equals, 7 seconds?
Which expression should Divya use to solve the problem?

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:
TypeCommon promptAppropriate expression
Displacement"What is the particle's displacement between... and..." or "What is the change in the particle's position between... and..."integral, start subscript, a, end subscript, start superscript, b, end superscript, v, left parenthesis, t, right parenthesis, d, t
Total distance"What is the total distance the particle has traveled between... and..."integral, start subscript, a, end subscript, start superscript, b, end superscript, \mid, v, left parenthesis, t, right parenthesis, \mid, d, t
Actual position"What is the particle's position at..."C, plus, integral, start subscript, a, end subscript, start superscript, b, end superscript, v, left parenthesis, t, right parenthesis, d, t where C is the initial condition

## Want to join the conversation?

• How do I know at what time a particle returns to the origin? • 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.
• 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? • • 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. 