Question

1. a motorcycle is moving in a straight line on a road. the distance traveled by the bike at various times is shown in the table below.

time (seconds) | 0 | 5 | 10 | 15
distance (feet) | 0 | 40 | 80 | 120
the dependent variable of this relation is
max value min value
the relation is increasing over the interval
what is the average rate of change interval x = 5 to x = 15
find the value of f(5) when f(x) = 80, x =
write an equation to express this relation

Explanation:

Step1: Identify Dependent Variable

The dependent variable depends on the independent variable (time here). Distance depends on time, so dependent variable is distance.

Step2: Find Max and Min Distance

Distance values: 0, 40, 80, 120. Max is 120, Min is 0.

Step3: Increasing Interval

Time increases from 0 to 15, distance increases. Interval: \(0 \leq t \leq 15\) (or \( [0,15] \)).

Step4: Average Rate of Change (5 to 15)

Formula: \(\frac{\Delta y}{\Delta x}=\frac{120 - 40}{15 - 5}=\frac{80}{10}=8\).

Step5: Find \(f(5)\) and \(x\) for \(f(x)=80\)

From table, \(f(5)=40\); \(f(x)=80\) when \(x = 10\).

Step6: Equation of Relation

Slope \(m=\frac{40 - 0}{5 - 0}=8\), y - intercept \(b = 0\). Equation: \(y = 8t\) (or \(d = 8t\)).

Answer:

• Dependent variable: Distance (feet)
• Max value: 120
• Min value: 0
• Increasing interval: \(0\leq t\leq15\) (or \([0,15]\))
• Average rate of change (5 to 15): 8
• \(f(5)\): 40; \(x\) when \(f(x)=80\): 10
• Equation: \(d = 8t\) (or \(y = 8x\))