Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 9 - Differential Equations

Problem 9.1.44a

Chapter 9, Problem 9.1.44a

43–44. Motion in a gravitational field: An object is fired vertically upward with initial velocity v(0)=v₀ from initial position s(0)=s₀.
a. For the following values of v₀ and s₀, find the position and velocity functions for all times at which the object is above the ground (s = 0).
v₀ = 49 m/s, s₀ = 60 m

Verified step by step guidance

Identify the physical model: The motion of the object under gravity can be described by the second-order differential equation for position \(s(t)\(: \[\frac{d^2 s}{dt^2} = -g,\] where \)g = 9.8 \ \text{m/s}^2\) is the acceleration due to gravity acting downward.

Integrate the acceleration to find the velocity function \(v(t)\): Since \(v(t) = \frac{ds}{dt}\), integrate the acceleration once to get \[v(t) = -g t + C_1,\] where \(C_1\) is a constant determined by the initial velocity condition.

Apply the initial velocity condition \(v(0) = v_0 = 49 \ \text{m/s}\) to find \(C_1\): Substitute \(t=0\) into the velocity function to get \[v(0) = C_1 = 49,\] so \[v(t) = -9.8 t + 49.\]

Integrate the velocity function to find the position function \(s(t)\(: \[s(t) = \int v(t) dt = \int (-9.8 t + 49) dt = -4.9 t^2 + 49 t + C_2,\] where \)C_2\) is a constant determined by the initial position.

Apply the initial position condition \(s(0) = s_0 = 60 \ \text{m}\) to find \(C_2\): Substitute \(t=0\( into the position function to get \[s(0) = C_2 = 60,\] so \[s(t) = -4.9 t^2 + 49 t + 60.\] The object is above the ground for all \)t\) such that \(s(t) > 0\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Kinematic Equations for Motion Under Constant Acceleration

These equations describe the position and velocity of an object moving with constant acceleration, such as gravity. Position is given by s(t) = s₀ + v₀t + (1/2)at², and velocity by v(t) = v₀ + at, where a is acceleration due to gravity (usually -9.8 m/s²). They allow calculation of motion parameters at any time t.

Initial Conditions in Differential Equations

Initial conditions specify the starting position and velocity of the object, here s(0) = s₀ and v(0) = v₀. These values are essential to uniquely determine the position and velocity functions over time when solving the motion equations.

Determining the Time Interval When the Object is Above Ground

To find when the object is above ground, solve s(t) > 0 using the position function. This involves finding the roots of the quadratic equation s(t) = 0, which mark the times the object is at ground level. The object is above ground between these roots.

Recommended video:

06:47

Derivatives Applied To Velocity Example 1

Related Practice

Textbook Question

17–20. Increasing and decreasing solutions Consider the following differential equations. A detailed direction field is not needed.

a. Find the solutions that are constant, for all t ≥ 0 (the equilibrium solutions).

y'(t) = (y−2)(y+1)

views

Textbook Question

33–36. {Use of Tech} Computing Euler approximations Use a calculator or computer program to carry out the following steps.

a. Approximate the value of y(T) using Euler’s method with the given time step on the interval [0,T].

y′(t) = t/y, y(0) = 4; Δt = 0.1, T = 2; y(t) = √(t² + 16)

Textbook Question

42–43. Implicit solutions for separable equations For the following separable equations, carry out the indicated analysis.

a. Find the general solution of the equation.

y'(t) = t²/(y² + 1); y(−1) = 1, y(0) = 0, y(−1) = −1

views

Textbook Question

{Use of Tech} Endowment model An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem B′(t)=rB−m, for t≥0, with B(0)=B0. The constant r>0 reflects the annual interest rate, m>0 is the annual rate of withdrawal, B0 is the initial balance in the account, and t is measured in years.

a. Solve the initial value problem with r=0.05, m=\(1000/year, and B0=\)15,000 Does the balance in the account increase or decrease?

views

Textbook Question

27–30. Predator-prey models Consider the following pairs of differential equations that model a predator-prey system with populations x and y. In each case, carry out the following steps.

a. Identify which equation corresponds to the predator and which corresponds to the prey.

x′(t) = −3x + xy, y′(t) = 2y − xy

Textbook Question

38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.

a. Find the equilibrium solutions.

y′(t) = 6 - 2y

views