Chapter 8.1, Problem 20E

a.

To determine

To find where the particle is at the end of the trip.

Answer to Problem 20E

The Position of the particle at the end of the trip is 5.

Explanation of Solution

Given information:

The particle is moving on x-axis and starts at $x=2$ when $t=0$ .

Calculation:

As displacement is equals to area under velocity time curve and displacement is a vector quantity therefore area enclosed when velocity is negative is negative displacement.

  $\begin{array}{l}\text{position at}\left(t=10\right)=\text{position at}\left(t=0\right)+\text{displacement from }\left(t=0\right)\text{to}\left(t=10\right)\\ \text{position at}\left(t=10\right)=2+\left(\frac{1}{2}\cdot 3\cdot 2\right)-\left(\frac{1}{2}\cdot 3\cdot 1+3\cdot 3+\frac{1}{2}\cdot 3\cdot 1\right)+\left(\frac{1}{2}\cdot 3\cdot 3\right)\\ =2+3-12+4.5\\ =-2.5\end{array}$

Therefore,

The Position of the particle at the end of the trip is $-2.5$ .

b.

To determine

To find the total distance travelled at the end of the trip.

Answer to Problem 20E

Total distance travelled at the end of the trip is 19.5 meter.

Explanation of Solution

Given information:

The particle is moving on x-axis and starts at $x=2$ when $t=0$ .

Calculation:

Total distance(S) travelled is equals to total area under velocity time curve.

  $\begin{array}{l}S=\left(\frac{1}{2}\cdot 3\cdot 2\right)+\left(\frac{1}{2}\cdot 3\cdot 1+3\cdot 3+\frac{1}{2}\cdot 3\cdot 1\right)+\left(\frac{1}{2}\cdot 3\cdot 3\right)\\ =19.5\end{array}$

Therefore,

Total distance travelled is 19.5 meter.