Question

Piecewise velocity The velocity of a (fast) automobile on a straight highway is given by the function﻿v(t)={3t if 0≤t\u003c2060 if 20≤t\u003c45240−4t if t≥45
v(t)= \begin{cases}3 t & 	ext { if } 0 \leq t\u003c20 \ 60 & 	ext { if } 20 \leq t\u003c45 \ 240-4 t & 	ext { if } t \geq 45\end{cases}
v(t)=⎩⎨⎧​3t60240−4t​ if 0≤t\u003c20 if 20≤t\u003c45 if t≥45​﻿ , where is measured in seconds and v has units of m/s. c. What is the distance traveled by the automobile in the first 60 s?

Accepted Answer

Understand that the distance traveled by the automobile over a time interval is the integral of the velocity function over that interval. Since velocity is given piecewise, we will integrate each piece over its respective time interval and then sum the results. Set up the integral for the first interval where $$v(t) = 3t$$ for $$0 \leq t \u003c 20. $$The distance traveled in this interval is $$\$$int_0$$^{20} 3t \, dt. $$Set up the integral for the second interval where $$v(t) = 60$$ for $$20 \leq t \u003c 45. $$The distance traveled in this interval is $$\int_{20}^{45} 60 \, dt. $$Set up the integral for the third interval where $$v(t) = 240 - 4t$$ for $$t \geq 45. $$Since we want the distance up to 60 seconds, integrate from 45 to 60: $$\int_{45}^{60} (240 - 4t) \, dt. $$Calculate each integral separately and then add the three results together to find the total distance traveled in the first 60 seconds.

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Key Concepts

Piecewise Functions

Distance from Velocity via Integration

Handling Piecewise Integration