Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.1.65c

Chapter 6, Problem 6.1.65c

Bike race Theo and Sasha start at the same place on a straight road, riding bikes with the following velocities (measured in mi/hr). Assume t is measured in hours.
Theo: vT(t)=10, for t≥0
Sasha: vS(t)=15t, for 0≤t≤1, and vS(t)=15, for t>1

c. If the riders ride for 2 hr, who rides farther? Interpret your answer geometrically using the graphs of part (a).

Verified step by step guidance

Understand that the distance each rider travels is the integral of their velocity function over the time interval from 0 to 2 hours. This is because distance is the area under the velocity-time graph.

For Theo, whose velocity is constant at \(v_T(t) = 10\) mi/hr, calculate the distance by integrating the constant velocity over 2 hours: \(\int_0^2 10 \, dt\).

For Sasha, whose velocity changes, split the integral into two parts: from 0 to 1 hour where \(v_S(t) = 15t\), and from 1 to 2 hours where \(v_S(t) = 15\). So, calculate \(\int_0^1 15t \, dt + \int_1^2 15 \, dt\).

Evaluate both integrals (without computing the final numerical values here) to find the total distance each rider covers in 2 hours.

Compare the two distances to determine who rides farther. Geometrically, this corresponds to comparing the areas under each velocity curve on the graph from 0 to 2 hours.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Velocity as a Function of Time

Velocity describes how fast an object moves and in which direction, often expressed as a function of time. Understanding velocity functions allows us to analyze how speed changes over time, such as Theo's constant velocity and Sasha's piecewise velocity that increases then remains constant.

Distance Traveled as the Integral of Velocity

The total distance traveled over a time interval is found by integrating the velocity function over that interval. This means calculating the area under the velocity-time graph, which represents the accumulation of movement over time.

Interpreting Graphs of Piecewise Functions

Piecewise functions have different expressions over different intervals. Interpreting their graphs involves understanding how the function changes shape, such as Sasha's velocity increasing linearly then becoming constant, which affects the area under the curve and thus the total distance.

Recommended video:

05:36

Piecewise Functions

Related Practice

Textbook Question

Use the region R that is bounded by the graphs of y=1+√x,x=4, and y=1 complete the exercises.

Region R is revolved about the y-axis to form a solid of revolution whose cross sections are washers.

d. Write an integral for the volume of the solid.

views

Textbook Question

9–10. Velocity graphs The figures show velocity functions for motion along a line. Assume the motion begins with an initial position of s(0)=0. Determine the following.

d. A piecewise function for s(t)

views

Textbook Question

Cycling distance A cyclist rides down a long straight road with a velocity (in m/min) given by v(t) = 400−20t, for 0≤t≤10, where t is measured in minutes.

c. How far has the cyclist traveled when her velocity is 250 m/min?

views

Textbook Question

Acceleration A drag racer accelerates at a(t)=88 ft/s². Assume v(0)=0, s(0)=0, and t is measured in seconds.

d. How long does it take the racer to travel 300 ft?

views

Textbook Question

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.

c. Find the distance traveled over the given interval.

v(t) = 3t²−6t on [0, 3]

views

Textbook Question

Flying into a headwind The velocity (in mi/hr) of an airplane flying into a headwind is given by v(t) = 30(16−t²), for 0≤t≤3. Assume s(0)=0 and t is measured in hours.

c. How far has the airplane traveled at the instant its velocity reaches 400 mi/hr?

views