Chapter 15.2, Problem 46E

(a)

To determine

To find: The car’s velocity at $t=12$ .

Answer to Problem 46E

The car’s velocity at $t=12$ is $81\text{ft/s}$ .

Explanation of Solution

Given information:

The given equation is $v\left(t\right)=15+4t+\frac{1}{8}{t}^2$ .

Calculation:

The car’s velocity at $t=12$ .

  $\begin{array}{c}v\left(12\right)=15+4\left(12\right)+\frac{1}{8}{\left(12\right)}^2\\ =15+48+18\\ =81\end{array}$

Therefore, the car’s velocity at $t=12$ is $81\text{ft/s}$ .

(b)

To determine

To find: The car’s acceleration at $t=12$ .

Answer to Problem 46E

The car’s acceleration at $t=12$ is $7{\text{ft/s}}^2$ .

Explanation of Solution

Given information:

The given equation is $v\left(t\right)=15+4t+\frac{1}{8}{t}^2$ .

Calculation:

The car’s acceleration is denoted by $a\left(t\right)$ .

  $a\left(t\right)=\frac{dv}{dt}$

  $\begin{array}{c}a\left(t\right)=v\text{'}\left(t\right)\\ =4+\frac{t}{4}\end{array}$

Calculate the car’s acceleration at $t=12$ .

  $\begin{array}{c}a\left(12\right)=4+\frac{12}{4}\\ =7\end{array}$

Therefore, the car’s acceleration at $t=12$ is $7{\text{ft/s}}^2$ .

(c)

To determine

To find: The interpretation of part b in words.

Answer to Problem 46E

The car’s acceleration at $t=12$ is $7{\text{ft/s}}^2$ .

Explanation of Solution

Given information:

The given equation is $v\left(t\right)=15+4t+\frac{1}{8}{t}^2$ .

Calculation:

The car’s acceleration at $t=12$ is $7{\text{ft/s}}^2$ .

Therefore, the car’s acceleration at $t=12$ is $7{\text{ft/s}}^2$ .

(d)

To determine

To find: The equation for $s\left(t\right)$ .

Answer to Problem 46E

The equation for $s\left(t\right)$ is $s\left(t\right)=15t+t^2+\frac{t^3}{24}$ .

Explanation of Solution

Given information:

The given equation is $v\left(t\right)=15+4t+\frac{1}{8}{t}^2$ .

Calculation:

The car’s velocity is.

  $v\left(t\right)=\frac{ds}{dt}$

Here, $s$ is the displacement of car at time $t$ .

  $v\left(t\right)=s\text{'}\left(t\right)$

  $s\left(t\right)$ is anti-derivative of $v\left(t\right)=15+4t+\frac{1}{8}{t}^2$ .

  $s\left(t\right)=15t+t^2+\frac{t^3}{24}+c$

  $c$ is an arbitrary constant.

At $t=0$ , $s\left(t\right)=0$ .

  $\begin{array}{c}0=15\left(0\right)+{\left(0\right)}^2+\frac{{\left(0\right)}^3}{24}+c\\ c=0\end{array}$

So, the equation for $s\left(t\right)$ is.

  $s\left(t\right)=15t+t^2+\frac{t^3}{24}$

Therefore, the equation for $s\left(t\right)$ is $s\left(t\right)=15t+t^2+\frac{t^3}{24}$ .

(e)

To determine

To find: The distance of car travelled in first $12\sec$ .

Answer to Problem 46E

The distance of car travelled in first $12\sec$ is $404\text{ft}$ .

Explanation of Solution

Given information:

The given equation is $v\left(t\right)=15+4t+\frac{1}{8}{t}^2$ .

Calculation:

The equation for $s\left(t\right)$ is.

  $s\left(t\right)=15t+t^2+\frac{t^3}{24}$

Calculate the distance of car travelled in first $12\sec$ .

  $\begin{array}{c}s\left(12\right)=15\left(12\right)+{\left(12\right)}^2+\frac{{\left(12\right)}^3}{24}\\ =180+144+72\\ =404\text{ft}\end{array}$

Therefore, the distance of car travelled in first $12\sec$ is $404\text{ft}$ .