Skip to main content
Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.2.47
Chapter 4, Problem 4.2.47

Finding Position from Velocity or Acceleration


Exercises 45–48 give the acceleration a=d²s/dt², initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t.


a = 9.8, v(0) = −3, s(0) = 0

Verified step by step guidance
1
Start by integrating the acceleration function a(t) = 9.8 with respect to time t to find the velocity function v(t). This involves finding the antiderivative of a constant function.
The antiderivative of a constant 9.8 is 9.8t. Therefore, v(t) = 9.8t + C, where C is the constant of integration.
Use the initial condition v(0) = -3 to solve for the constant C. Substitute t = 0 and v(0) = -3 into the velocity equation: -3 = 9.8(0) + C.
Solve for C to find that C = -3. Thus, the velocity function is v(t) = 9.8t - 3.
Integrate the velocity function v(t) = 9.8t - 3 with respect to time t to find the position function s(t). Use the initial condition s(0) = 0 to solve for the constant of integration in the position function.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

Key Concepts

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

Integration

Integration is the process of finding the antiderivative or the area under a curve. In this context, it is used to find the velocity function from the acceleration function by integrating acceleration with respect to time. This step is crucial for determining the velocity at any given time.
Recommended video:
05:04
Introduction to Indefinite Integrals

Initial Conditions

Initial conditions are values given at the start of a problem that help determine the specific solution to a differential equation. Here, the initial velocity v(0) = -3 and initial position s(0) = 0 are used to find the constants of integration when solving for velocity and position functions.
Recommended video:
05:03
Initial Value Problems

Position Function

The position function s(t) describes the location of an object at any time t. It is found by integrating the velocity function, which itself is derived from the acceleration function. The position function is essential for understanding the object's movement over time and is determined using integration and initial conditions.
Recommended video:
5:20
Relations and Functions
Related Practice
Textbook Question

6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 2x − 7, y(2) = 0

Textbook Question

Theory and Examples


Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀,  g > 0, with s in meters and t in seconds. Find the body’s maximum height.

Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(cscθ cotθ) / 2 dθ

Textbook Question

[Technology Exercises] When solving Exercises 14–30, you may need to use appropriate technology (such as a calculator or a computer).

26. Factoring a quartic Find the approximate values of r_1 through r_4 in the factorization

8x^4-14x^3-9x^2+11x-1=8(x-r_1)(x-r_2)(x-r_3)(x-r_4)

Textbook Question

Finding Critical Points


In Exercises 41–50, determine all critical points and all domain endpoints for each function.


y = x − 3x²ᐟ³