Constant Rate Math Example 1

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Example 1

easy
A car travels at a constant speed of 6060 km/h. Write the distance function d(t)d(t), find d(2.5)d(2.5), and compute the rate of change from t=1t=1 to t=3t=3.

Solution

  1. 1
    Constant rate 6060 km/h means distance == rate ×\times time: d(t)=60td(t) = 60t.
  2. 2
    Evaluate: d(2.5)=60×2.5=150d(2.5) = 60 \times 2.5 = 150 km.
  3. 3
    Average rate of change from t=1t=1 to t=3t=3: d(3)−d(1)3−1=180−602=1202=60\frac{d(3)-d(1)}{3-1} = \frac{180-60}{2} = \frac{120}{2} = 60 km/h. (Constant rate means average rate equals instantaneous rate.)

Answer

d(t)=60td(t)=60t; d(2.5)=150d(2.5)=150 km; rate =60=60 km/h
A constant rate of change produces a linear function. For any linear function f(x)=mx+bf(x)=mx+b, the average rate of change between any two points always equals the slope mm.

About Constant Rate

A constant rate of change means the output increases (or decreases) by the same fixed amount for every unit increase in the input — the hallmark of a linear function.

Learn more about Constant Rate →

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