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Difficulty: Very hardSlope of a Line

A toy car moves along a straight track. Its position in meters, ss, is plotted against time in seconds, tt, on a standard coordinate plane. At t=2t = 2 seconds, the car is at a position of −3-3 meters. From t=2t = 2 to t=6t = 6 seconds, the position changes at a constant rate of mm meters per second. From t=6t = 6 to t=8t = 8 seconds, the position changes at a constant rate of 2m+32m + 3 meters per second. If the average rate of change of the car's position over the entire interval from t=2t = 2 to t=8t = 8 seconds is 53\frac{5}{3} meters per second, what is the constant rate of change, in meters per second, from t=6t = 6 to t=8t = 8 seconds?

  1. A
    1/21/2
  2. B
    12/5
  3. C
    29/9
  4. 4Answer
  5. E
    19/4

Answer

The correct constant rate of change from t=6t = 6 to t=8t = 8 seconds is 44 meters per second.
The correct answer of 44 meters per second is found by setting up the displacement for each interval in terms of mm. The displacement during the first interval is 4m4m, and the displacement during the second interval is 2(2m+3)=4m+62(2m + 3) = 4m + 6. Adding these gives a total displacement of 8m+68m + 6 over a total time of 66 seconds. Dividing total displacement by total time yields the average rate of change, 8m+66=4m+33\frac{8m + 6}{6} = \frac{4m + 3}{3}. Setting this equal to the given average rate of 53\frac{5}{3} yields m=12m = \frac{1}{2}. Finally, substituting this back into the rate expression for the second interval, 2m+32m + 3, gives 2(1/2)+3=42(1/2) + 3 = 4.

Step-by-Step Solution

1
Express the displacement in each time interval using the rate of change (slope) formula: Δs=slope×Δt\Delta s = \text{slope} \times \Delta t.
For t=2t = 2 to t=6t = 6: Δs1=m×(6−2)=4m\Delta s_1 = m \times (6 - 2) = 4m. For t=6t = 6 to t=8t = 8: Δs2=(2m+3)×(8−6)=2(2m+3)=4m+6\Delta s_2 = (2m + 3) \times (8 - 6) = 2(2m + 3) = 4m + 6.
The constant rate of change in a position-time graph is the slope of the line, which relates time intervals to position displacements.
2
Calculate the total displacement over the entire interval from t=2t = 2 to t=8t = 8 seconds.
Total displacement Δstotal=Δs1+Δs2=4m+(4m+6)=8m+6\Delta s_{\text{total}} = \Delta s_1 + \Delta s_2 = 4m + (4m + 6) = 8m + 6.
The total displacement is the sum of the individual displacements over consecutive sub-intervals.
3
Set up the average rate of change equation using the total displacement and the total time elapsed (Δttotal=8−2=6\Delta t_{\text{total}} = 8 - 2 = 6 seconds).
Average rate of change = 8m+66=4m+33\frac{8m + 6}{6} = \frac{4m + 3}{3}.
The average rate of change is the net displacement divided by the total time elapsed.
4
Equate the expression for the average rate of change to the given value of 53\frac{5}{3} and solve for mm.
4m+33=53⇒4m+3=5⇒4m=2⇒m=12\frac{4m + 3}{3} = \frac{5}{3} \Rightarrow 4m + 3 = 5 \Rightarrow 4m = 2 \Rightarrow m = \frac{1}{2}.
Solving this linear equation yields the value of the parameter mm that satisfies the average rate condition.
5
Calculate the rate of change for the second interval from t=6t = 6 to t=8t = 8 seconds by substituting m=12m = \frac{1}{2} into 2m+32m + 3.
Rate of change = 2(12)+3=1+3=42\left(\frac{1}{2}\right) + 3 = 1 + 3 = 4.
The question asks for the rate of change during the second interval, which is defined in terms of mm as 2m+32m + 3.

Key Concept

Slope as a constant rate of change in a piecewise linear model
Estimated Time:3m 0s
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Topics

Slope of a Line

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