Chapter 3, Problem 23P

To determine

The distance travelled by the car between $t=0\text{s}$ to $t=16\text{s}$, the motion diagram showing the position of the car at $2\text{s}$ intervals and the graph of $x\left(t\right)$.

Answer to Problem 23P

The distance travelled by the car between $t=0\text{s}$ to $t=16\text{s}$ is $\underset{\_}{160\text{m}}$, figure $1$ shows the motion diagram showing the position of the car at $2\text{s}$ intervals and figure $2$ shows the graph of $x\left(t\right)$.

Explanation of Solution

Write the expression for the displacement from the velocity-time graph.

    $\text{displacement}=\text{ area under the graph}$        (I)

Write the expression for the area under the graph.

    $\text{area under the graph}=\text{area of the rectangle}\left(t=0-4\text{s}\right)+\text{area of the triangle}\left(t=4-12\text{s}\right)$        (II)

Write the expression for the area of rectangle ABCD.

    $A_{\text{Rectangle}}=\left(\text{AB}\right)\left(\text{BC}\right)$        (III)

Here, $A_{\text{Rectangle}}$ is the area of the rectangle, BC is the length, AB is the height of the rectangle.

Write the expression for the area of the triangle CDE

    $A_{\text{triangle}}=\frac{1}{2}\left(\text{DE}\right)\left(\text{CD}\right)$        (IV)

Here, $A_{\text{triangle}}$ is the area of triangle, DE is the base, CD is the height of triangle.

Use equation (III) and (IV) in (II) to solve for distance travelled by the car.

    $\text{area under the graph}=\left(\text{AB}\right)\left(\text{BC}\right)+\frac{1}{2}\left(\text{DE}\right)\left(\text{CD}\right)$        (V)

Write the expression for the distance travelled by an object.

    $\begin{array}{c}\text{distance}=\text{velocity}\times \text{time}\\ =\text{area under the graph}\end{array}$        (VI)

Use equation (VI) and (V) to solve for the position of the car at $2\text{s}$ intervals.

    $\begin{array}{c}x\left(t=0\text{s}\right)=\left(20\text{m}/\text{s}\right)\left(0\text{s}\right)\\ =0\\ x\left(t=2\text{s}\right)=\left(20\text{m}/\text{s}\right)\left(2\text{s}\right)\\ =40\text{m}\\ x\left(t=4\text{s}\right)=\left(20\text{m}/\text{s}\right)\left(4\text{s}\right)\\ =80\text{m}\\ x\left(t=6\text{s}\right)=\left(20\text{m}/\text{s}\right)\left(4\text{s}\right)+\left(15\text{m}/\text{s}\right)\left(2\text{s}\right)+\left(1/2\right)\left(2\text{s}\right)\left(5\text{m}/\text{s}\right)\\ =115\text{m}\end{array}$        (VII)

Similarly, the distance travelled by the car at $2\text{s}$ intervals of time using the idea of finding the area under the graph.

Figure $1$ shows the motion diagram showing the position of the car at $2\text{s}$ intervals.

Figure $2$ shows the graph of $x\left(t\right)$.

Conclusion:

Substitute $20\text{m}/\text{s}$ for AB, $4\text{s}$ for BC, $20\text{m}/\text{s}$ for CD, $8\text{s}$ for DE in equation (V) to find the distance travelled by the car between $t=0\text{s}$ to $t=16\text{s}$.

    $\begin{array}{c}\text{area under the graph}=\left(20\text{m}/\text{s}\right)\left(4\text{s}\right)+\frac{1}{2}\left(8\text{s}\right)\left(20\text{m}/\text{s}\right)\\ =160\text{m}\end{array}$

Therefore, the distance travelled by the car between $t=0\text{s}$ to $t=16\text{s}$ is $\underset{\_}{160\text{m}}$, figure $1$ shows the motion diagram showing the position of the car at $2\text{s}$ intervals and figure $2$ shows the graph of $x\left(t\right)$.