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Concept explainers
a.
To graph: the truck’s velocity v=ds/dt for
a.
![Check Mark](/static/check-mark.png)
Answer to Problem 26E
Explanation of Solution
Given information: The graph shows the position s of truck traveling on a highway. The Truck starts at t =0 and return 15 hours later at t =15.
Calculation:
The velocity function v (in dark blue) is the derivative of the position function is s. The derivative at point t gives the slope of the tangent line to the original function at point t.
Starting at t = 0. The tangent to s is horizontal so v=0. Following along the curve of s up to about t = 5, the slope of the tangent lines are increasing, so v increases in the same interval.
After t =5, the slope of the tangent lines starts to decrease until it gets horizontal again at t =10. So from t =5 to t = 10, v decreases and goes back to 0.
After t = 10, the slope of the tangent lines becomes increasingly negative all the way through t = 15. So after t = 10, v is increasingly negative.
Acceleration a (in red) is the derivative of velocity because the derivative is just the slope of the tangent lines. Before t = 5, the slope of the tangent to v are positive but decreasing to 0. So a is positive and decreasing to 0 at t = 5.
After t = 5. The slope of the tangent to v is negative and continues to decrease at the same rate.
So a is negative and continues in the same direction.
b.
To graph: ds/dt and
b.
![Check Mark](/static/check-mark.png)
Answer to Problem 26E
Explanation of Solution
Given information: The graph shows the position s of truck traveling on a highway. The Truck starts at t =0 and return 15 hours later at t =15.
Calculation:
Given s, so find the first and second derivative using power rule.
The graph of s, v=ds/dt and
For all of them the horizontal axis is t in hours, but for v the vertical axis is km per hours and for a if is km/
Chapter 3 Solutions
Calculus: Graphical, Numerical, Algebraic
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