Number and Operations>Representing Constant Rates of Change Including d = rt(TEKS.Math.7.4.A)
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Texas 7th Grade Math › Number and Operations>Representing Constant Rates of Change Including d = rt(TEKS.Math.7.4.A)
A train travels at a constant speed. At 2 hours it has gone 150 miles; at 5 hours it has gone 375 miles. Use $d = rt$.
What is the train's constant speed?
50 miles/hour
60 miles/hour
75 miles/hour
225 miles/hour
Explanation
A constant rate means the ratio distance/time stays the same. Compute $r = \frac{d}{t}$: $\frac{150}{2} = 75$ and $\frac{375}{5} = 75$. In $d = rt$, $r$ is 75, so the speed is 75 miles/hour.
Consider these four representations of distance over time.
Which representation shows a constant rate of change?
Table: time (h) 1, 2, 3; distance (mi) 52, 104, 156
Table: time (h) 1, 2, 3; distance (mi) 50, 110, 165
Equation: $d = 120t^2$
Verbal: A jogger speeds up, covering 0.9 mile in the first 10 minutes and 1.1 miles in the next 10 minutes
Explanation
A constant rate has a consistent ratio $\frac{d}{t}$. Choice A: $\frac{52}{1}=52$, $\frac{104}{2}=52$, $\frac{156}{3}=52$ (constant). B changes (50, 55, 55). C is quadratic ($t^2$) so the rate varies with $t$. D describes speeding up, not a constant rate.
A tank is being filled. Points from its volume–time graph are $(0, 5)$, $(2, 11)$, and $(6, 23)$, where time is in minutes and volume is in gallons.
What is the constant rate of change of the tank's volume?
1.5 gallons/minute
2 gallons/minute
2.5 gallons/minute
3 gallons/minute
Explanation
Constant rate equals the slope $\frac{\Delta V}{\Delta t}$. From $(0,5)$ to $(2,11)$: $\frac{11-5}{2-0}=\frac{6}{2}=3$. From $(2,11)$ to $(6,23)$: $\frac{23-11}{6-2}=\frac{12}{4}=3$. The rate is 3 gallons/minute.
A taxi charges according to $C = 4 + 2.50m$, where $C$ is the cost in dollars and $m$ is the number of miles ridden.
What is the constant rate of change and what does it represent?
4 dollars per mile; the starting fee
2.50 dollars per mile; the fare increases by 2.50 for each additional mile
2.50 dollars; the one-time starting fee
6.50 dollars per mile; the total cost per mile
Explanation
In a linear model, the constant rate of change is the coefficient of the variable: here $2.50$ is multiplied by $m$, so the rate is 2.50 dollars per mile. The 4 is the fixed starting fee (the $y$-intercept).
A cyclist's distance increases steadily over time as shown in the table: time (h): 1, 2, 3, 4; distance (mi): 18, 36, 54, 72.
Which equation best models this relationship using $d = rt$?
$d = 18t$
$d = t + 18$
$d = 18 + t^2$
$d = 18t + t$
Explanation
A constant rate has a constant ratio $\frac{d}{t}$. Here $\frac{18}{1}=\frac{36}{2}=\frac{54}{3}=\frac{72}{4}=18$, so $r=18$. Using $d = rt$, the matching equation is $d = 18t$. The other options are not constant-rate models for this table.