Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

10. Physics Applications of Integrals

Kinematics

Calculus

10. Physics Applications of Integrals

Kinematics

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1:39 minutes

Problem 6.1.19a

Textbook Question

17–22. Position from velocity Consider an object moving along a line with the given velocity v and initial position.

a. Determine the position function, for t≥0, using the antiderivative method

v(t) = 6−2t on [0, 5]; s(0)=0

Verified step by step guidance

Identify the given velocity function: $v(t) = 6 - 2t$ and the initial position $s(0) = 0$.

Recall that the position function $s(t)$ is the antiderivative (indefinite integral) of the velocity function $v(t)$, so write $s(t) = \int v(t) \, dt + C$.

Integrate the velocity function: compute $\int (6 - 2t) \, dt$ by integrating each term separately.

After integration, express $s(t)$ as $s(t) = 6t - t^2 + C$, where $C$ is the constant of integration.

Use the initial condition $s(0) = 0$ to solve for $C$ by substituting $t=0$ into $s(t)$ and setting it equal to zero.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Velocity and Position Relationship

Velocity is the rate of change of position with respect to time. The position function can be found by integrating the velocity function over time, which accumulates the changes in position from the initial point.

Antiderivative (Indefinite Integral)

The antiderivative of a function is another function whose derivative is the original function. Finding the antiderivative of velocity gives the position function up to a constant, which can be determined using initial conditions.

Initial Conditions and Constants of Integration

When integrating to find position from velocity, an unknown constant appears. The initial position value (e.g., s(0) = 0) is used to solve for this constant, ensuring the position function accurately reflects the object's starting point.

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Textbook Question

Flying into a headwind The velocity (in mi/hr) of an airplane flying into a headwind is given by v(t) = 30(16−t²), for 0≤t≤3. Assume s(0)=0 and t is measured in hours.

c. How far has the airplane traveled at the instant its velocity reaches 400 mi/hr?

Textbook Question

Cycling distance A cyclist rides down a long straight road with a velocity (in m/min) given by v(t) = 400−20t, for 0≤t≤10, where t is measured in minutes.

c. How far has the cyclist traveled when her velocity is 250 m/min?

Textbook Question

Cycling distance A cyclist rides down a long straight road with a velocity (in m/min) given by v(t) = 400−20t, for 0≤t≤10, where t is measured in minutes.

b. How far does the cyclist travel in the first 10 min?

Textbook Question

17–22. Position from velocity Consider an object moving along a line with the given velocity v and initial position.

a. Determine the position function, for t≥0, using the antiderivative method

v(t) = 9−t² on [0, 4]; s(0)=−2

Textbook Question

17–22. Position from velocity Consider an object moving along a line with the given velocity v and initial position.

a. Determine the position function, for t≥0, using the antiderivative method

v(t) = −t³+3t²−2t on [0, 3]; s(0)=4

Textbook Question

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.

b. Find the displacement over the given interval.

v(t) = 50e^−2t on [0, 4]

Textbook Question

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.

a. Determine when the motion is in the positive direction and when it is in the negative direction.

v(t) = 50e^−2t on [0, 4]

Textbook Question

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.

c. Find the distance traveled over the given interval.

v(t) = 4t³ - 24t²+20t on [0, 5]

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