Chapter 4.8, Problem 42E

To determine

To calculate:The particle position after $t$ seconds when acceleration function of particle is $a\left(t\right)=5+4t-2t^2$ ,initial velocity is $v\left(0\right)=3m/s$ and initial displacement is $s\left(0\right)=10\text{}m$ .

Answer to Problem 42E

The position function of particle after $t$ seconds is $s\left(t\right)=5\frac{t^2}{2}+2\frac{t^3}{3}-\frac{t^4}{6}+3t+10$ .

Explanation of Solution

Given information:

The acceleration function of particle given as:

  $a\left(t\right)=5+4t-2t^2$ ,initial velocity is $v\left(0\right)=3m/s$ and initial displacement is $s\left(0\right)=10\text{}m$ .

Formula used:

The object has position function $s\left(t\right)=f\left(t\right)$ then the velocity function is $v\left(t\right)=s^{\text{'}}\left(t\right)$

The acceleration function is $a\left(t\right)=v^{\text{'}}\left(t\right)$ .

Calculation:

Consider the acceleration function,

  $a\left(t\right)=5+4t-2t^2$

For the position function firstly calculate the velocity function,

As, velocity function is an antiderivative of acceleration.

Therefore, integrate the equation $a\left(t\right)=v^{\text{'}}\left(t\right)$

  $\begin{array}{c}{\displaystyle \int a\left(t\right)}dt={\displaystyle \int v^{\text{'}}\left(t\right)}dt\\ {\displaystyle \int 5+4t-2t^2}dt=v\left(t\right)\\ 5t+4\frac{t^2}{2}-2\frac{t^3}{3}+C=v\left(t\right)\end{array}$

As, the position function is an antiderivative of the velocity function.

Therefore, integrate the equation $v\left(t\right)=s^{\text{'}}\left(t\right)$

  $\begin{array}{c}{\displaystyle \int v\left(t\right)}dt={\displaystyle \int s^{\text{'}}\left(t\right)}dt\\ {\displaystyle \int 5t+4\frac{t^2}{2}-2\frac{t^3}{3}+C}dt=s\left(t\right)\\ 5\frac{t^2}{2}+4\frac{t^3}{6}-2\frac{t^4}{12}+Ct+D=s\left(t\right)\\ 5\frac{t^2}{2}+2\frac{t^3}{3}-\frac{t^4}{6}+Ct+D=s\left(t\right)\end{array}$

Apply the initial velocity condition $v\left(0\right)=3m/s$ in velocity function $5t+4\frac{t^2}{2}-2\frac{t^3}{3}+C=v\left(t\right)$

  $\begin{array}{c}v\left(0\right)=5\left(0\right)+4\frac{{\left(0\right)}^2}{2}-2\frac{{\left(0\right)}^3}{3}+C\\ 3=C\end{array}$

Put the value of $C=3$ in position function,

  $s\left(t\right)=5\frac{t^2}{2}+2\frac{t^3}{3}-\frac{t^4}{6}+3t+D$

Apply the initial displacement $s\left(0\right)=10\text{}m$ in position function,

  $\begin{array}{c}s\left(0\right)=5\frac{0^2}{2}+2\frac{0^3}{3}-\frac{0^4}{6}+3\cdot 0+D\\ 10=D\end{array}$

Put the value of $D=10$ in position function,

  $s\left(t\right)=5\frac{t^2}{2}+2\frac{t^3}{3}-\frac{t^4}{6}+3t+10$

Thus, the position function of particle after $t$ seconds is $s\left(t\right)=5\frac{t^2}{2}+2\frac{t^3}{3}-\frac{t^4}{6}+3t+10$