Chapter 3.4, Problem 44E

To determine

To select:the values of time from given option when particle is moving in left direction.

Answer to Problem 44E

Correct option that is time when particle moves in left is E that is $t=4$ .

Explanation of Solution

Given information:

1. Position function of the particle in terms of time $t$ is as follows

$s\left(t\right)=2+7t-t^2$

2. five given options of time are as follows

1. $t=0$ .
2. $t=1$
3. $t=2$
4. $t=\frac{7}{2}$
5. $t=4$

Formula used:

Velocity of a particle at any time $t$ that is $v\left(t\right)$ having position function $s\left(t\right)$ is as follows.

  $v\left(t\right)=\frac{d}{dt}s\left(t\right)=s^{\prime }\left(t\right)$

Calculation:

Position function of particle is $s\left(t\right)=2+7t-t^2$

Evaluate velocity of particle at any time $t$ .

  $v\left(t\right)=\frac{d}{dt}s\left(t\right)$

  $=\frac{d}{dt}\left(2+7t-t^2\right)$

  $=\frac{d}{dt}2+7\frac{d}{dt}t-\frac{d}{dt}{t}^2$

  $=0+7-2t$

  $v\left(t\right)=7-2t$

When particle moves to left velocity will be negative that is $v\left(t\right)<0$ .

Evaluate value of $t$ for following inequality

  $v\left(t\right)<0$

  $7-2t<0$

Add $2t$ to both sides of inequality

  $7-2t+2t<0+2t$

  $7<2t$

Divide both sides by 2.

  $\frac{7}{2}<\frac{2t}{2}$

  $\frac{7}{2}<t$

  $t>\frac{7}{2}$

This means for value of $t$ greater that $\frac{7}{2}$ , particle moves in left direction and out of given option there is only one value which is greater than $\frac{7}{2}$ that is $4$ .

Hence correct option is E that is $\underset{\_}{t=4}$ .