Chapter 6, Problem 49RE

$\text{(a)}$

To determine

To find: The velocity as a function of $t$ .

Answer to Problem 49RE

The answer: $v=2t+3t^2+4$

Explanation of Solution

Given information:

The acceleration of a particle moving along a coordinate line is:

  $\frac{d^{2}s}{d{t}^2}=2+6t\text{m}/{\text{sec}}^2$

Calculation:

The position function's derivative is the velocity:

  $\begin{array}{c}v(t)=\frac{ds}{dt}\\ \Rightarrow \frac{d^{2}s}{d{t}^2}=2+6t\\ \to \frac{d}{dt}\left(\frac{ds}{dt}\right)=2+6t\\ \to \frac{d}{dt}v(t)=2+6t\\ \to \frac{dv}{dt}=2+6t\end{array}$

Separable the variables and anti differentiate as follows.

  $\begin{array}{c}dv=(2+6t)dt\\ {{\displaystyle \to \int }}^{\text{}}dv={{\displaystyle \int }}^{\text{}}2dt+{{\displaystyle \int }}^{\text{}}6tdt\\ {{\displaystyle \to \int }}^{\text{}}dv=2{{\displaystyle \int }}^{\text{}}dt+6{{\displaystyle \int }}^{\text{}}tdt\\ \to v=2t+6\cdot \frac{t^{1+1}}{1+1}+C\\ \to v=2t+3t^2+C\end{array}$

Then apply the initial condition to find $C$:

Put,

  $\begin{array}{c}v=4\\ t=0\end{array}$

Then simplify,

  $\begin{array}{c}4=2\cdot 0+3\cdot 0^2+C\\ \to 4=0+0+C\\ \to C=4\\ v=2t+3t^2+4\end{array}$

Therefore, the required velocity as a function of $t$ is $v=2t+3t^2+4$ .

$\text{(b)}$

To determine

To find: How far does the particle go from $t=0$ to

  $t=1$ in the first second of its journey.

Answer to Problem 49RE

The answer: $6\text{ meters }$

Explanation of Solution

Given information:

The acceleration of a particle moving along a coordinate line is:

  $\frac{d^{2}s}{d{t}^2}=2+6t\text{m}/{\text{sec}}^2$

Calculation:

  $\begin{array}{c}dv=(2+6t)dt\\ {{\displaystyle \to \int }}^{\text{}}dv={{\displaystyle \int }}^{\text{}}2dt+{{\displaystyle \int }}^{\text{}}6tdt\\ {{\displaystyle \to \int }}^{\text{}}dv=2{{\displaystyle \int }}^{\text{}}dt+6{{\displaystyle \int }}^{\text{}}tdt\\ \to v=2t+6\cdot \frac{t^{1+1}}{1+1}+C\\ \to v=2t+3t^2+C\end{array}$

Then apply the initial condition to find $C$:

Put,

  $\begin{array}{c}v=4\\ t=0\end{array}$

Then simplify,

  $\begin{array}{c}4=2\cdot 0+3\cdot 0^2+C\\ \to 4=0+0+C\\ \to C=4\\ v=2t+3t^2+4\end{array}$

Separable the variables and anti differentiate as follows.

  $\begin{array}{c}ds=\left(2t+3t^2+4\right)dt\\ \to {\displaystyle \int d}s={\displaystyle \int \left(2t+3t^2+4\right)}dt\\ \to {\displaystyle \int d}s={\displaystyle \int 2}tdt+{\displaystyle \int 3}{t}^{2}dt+{\displaystyle \int 4}dt\\ \to {\displaystyle \int d}s=2{\displaystyle \int t}dt+3{\displaystyle \int t^2}dt+4{\displaystyle \int d}t\\ \to s=2\cdot \frac{t^{1+1}}{1+1}+3\cdot \frac{t^{2+1}}{2+1}+4t+C\\ \to s=t^2+t^3+4t+C\end{array}$

The particle moves during the first second of its journey:

  $\begin{array}{c}s(1)-s(0)=\left(1^2+1^3+4\cdot 1+C\right)-\left(0^2+0^3+4\cdot 0+C\right)\\ =1+1+4+C-C\\ =6meters\end{array}$

Therefore, the required $6\text{ meters }$ far does the particle go from $t=0$ to

  $t=1$ in the first second of its journey.