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 $s(t) > 0$.

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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.

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