Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 9 - Differential Equations

Problema 9.1.44a

Capitolo 9, Problema 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

Guida verificata passo dopo passo

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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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.

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08:14

Using The Acceleration Function

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.

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04:00

Solutions to Basic Differential 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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06:47

Derivatives Applied To Velocity Example 1

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