Chapter 4.8, Problem 45E

To determine

To show: The equation ${\left[v\left(t\right)\right]}^2=v_0^2-19.6\left[s\left(t\right)-s_0\right]$.

Answer to Problem 45E

The proof of statement ${\left[v\left(t\right)\right]}^2=v_0^2-19.6\left[s\left(t\right)-s_0\right]$ for object that is projected upward is shown.

Explanation of Solution

Given data:

Initial velocity is $v_0$, and initial displacement is $s_0$.

Formula Used:

Write the expression for acceleration function $a\left(t\right)$.

$a\left(t\right)=v^{\prime }\left(t\right)$ (1)

Here,

$v^{\prime }\left(t\right)$ is first derivative of velocity function $v\left(t\right)$.

Write the expression for velocity function $v\left(t\right)$.

$v\left(t\right)=s^{\prime }\left(t\right)$ (2)

Here,

$s^{\prime }\left(t\right)$ is first derivative of displacement function $s\left(t\right)$.

Antiderivative of $t$ is $\frac{1}{2}{t}^2$ and 1 is $t$.

Calculation:

Initial velocity is $v_0$, that is $v\left(0\right)=v_0$ and the initial displacement is $s_0$, that is $s\left(0\right)=s_0$.

The motion of stone is close to ground, so the motion is considered as gravitational constant $\left(g\right)$, which is $9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$.

Write the expression for acceleration function $\left(a\left(t\right)\right)$.

$a\left(t\right)=-9.8$

Substitute $-9.8$ for $a\left(t\right)$ in equation (1),

$v^{\prime }\left(t\right)=-9.8$

Antiderivate the expression with respect to t,

$v\left(t\right)=-9.8t+C$ (3)

Here,

C is arbitrary constant.

Substitute 0 for t in equations (3),

$\begin{array}{c}v\left(0\right)=-9.8\left(0\right)+C\\ =C\end{array}$

Substitute $v_0$ for $v\left(0\right)$,

$C=v_0$

Substitute $v_0$ for C in equation (3),

$v\left(t\right)=-9.8t+v_0$ (4)

Substitute $-9.8t+v_0$ for $v\left(t\right)$ in equation (2),

$s^{\prime }\left(t\right)=-9.8t+v_0$

Antiderivate the expression with respect to t,

$s\left(t\right)=-9.8\left(\frac{1}{2}{t}^2\right)+v_{0}t+D$

$s\left(t\right)=-4.9t^2+v_{0}t+D$ (5)

Here,

D is arbitrary constant.

Substitute 0 for t in equation (5),

$\begin{array}{c}s\left(0\right)=-4.9\left(0^2\right)+v_0\left(0\right)+D\\ =0+D\\ =D\end{array}$

Substitute $s_0$ for $s\left(0\right)$,

$D=s_0$

Substitute $s_0$ for D in equation (5),

$s\left(t\right)=-4.9t^2+v_{0}t+s_0$

Apply square on both sides of equation (4).

$\begin{array}{c}{\left[v\left(t\right)\right]}^2={\left(-9.8t+v_0\right)}^2\\ ={\left(v_0\right)}^2+{\left(9.8t\right)}^2-2\left(9.8t\right)v_0\\ =v_0^2+96.04t^2-19.6v_0\end{array}$

Add and subtract the term $s_0$ on RHS of expression.

$\begin{array}{c}{\left[v\left(t\right)\right]}^2=v_0^2+96.04t^2-19.6v_0-s_0+s_0\\ =v_0^2-19.6\left[\left(-4.9t^2+v_0+s_0\right)-s_0\right]\end{array}$

Substitute $s\left(t\right)$ for $\left(-4.9t^2+v_0+s_0\right)$,

${\left[v\left(t\right)\right]}^2=v_0^2-19.6\left[s\left(t\right)-s_0\right]$

Thus, the proof of statement ${\left[v\left(t\right)\right]}^2=v_0^2-19.6\left[s\left(t\right)-s_0\right]$ for object that is projected upward is shown.