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Q28E

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Essential Calculus: Early Transcendentals
Found in: Page 369
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Illustration

Short Answer

Two cars, A and B, start side by side and accelerate from rest. The figure shows the graphs of their velocity functions.

(a) Which car is ahead after one minute? Explain.

(b) What is the meaning of the area of the shaded region?

(c) Which car is ahead after two minutes? Explain.

(d) Estimate the time at which the cars are again side by side.

Car \(A\) is ahead of \({\mathop{\rm car}\nolimits} B\) after one minute duration.

The area of the shaded region is \({s_A}(1) - {s_B}(1)\).

The car \(A\) is ahead of car \(B\) after two-minute duration.

At \(2.2\;{\rm{min}}\), the cars travel side by side.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Area under curve.

The general expression for the area under curve between \(t = 0\) and \(t = x\) is shown below:

\(\int_0^x v (t)dt = s(x)(1)\)

The car that is in the lead after one minute duration.

a)

The general expression for the area under curve between \(t = 0\) and \(t = x\) is shown below:

\(\int_0^x v (t)dt = s(x)(1)\)

Here, the velocity is \(v(t)\) and the displacement is \(s(x)\).

Rearrange Equation (1) for the area under curve \(A\) as shown below:

\(\int_0^x {{v_A}} (t)dt = {s_A}(x)\)

Here, the velocity of car \(A\) is \({v_A}(t)\) and the displacement of \({\mathop{\rm car}\nolimits} A\) is \({s_A}(x)\).

Rearrange Equation (1) for the area under curve \(B\) as shown below:

\(\int_0^x {{v_B}} (t)dt = {s_B}(x)\)

Here, the velocity of \({\mathop{\rm car}\nolimits} B\) is \({v_B}(t)\) and the displacement of \({\mathop{\rm car}\nolimits} B\) is \({s_B}(x)\).

Refer the graph of the velocity functions.

After 1 minute time duration, the area under the curve \(A\) is greater than the area under the curve \(B\). Hence, car \(A\) is ahead of car \(B\) after one minute duration.

Therefore, Car \(A\) is ahead of \({\mathop{\rm car}\nolimits} B\) after one minute duration.

The representation of the area of a shaded region.

b)

Refer the graph for the velocity functions.

- The area of the shaded region represents the distance of car \(A\) which is ahead of car \(B\) after 1 minute duration.

Let the distance travelled by car \(A\) after 1 minute be \({s_A}(1)\) and the distance by car \(B\) after 1 minute be \({s_B}(1)\). Thus, the distance travelled by car \(A\) is ahead of car \(B\) which is expressed as \({s_A}(1) - {s_B}(1)\). Therefore, the area of the shaded region represents \({s_A}(1) - {s_B}(1)\).

The car that is in the lead after two minutes.

c)

Refer the graph for the velocity functions.

The velocity of car \(A\) is greater than the car \(B\) after two minutes duration.

Therefore, the car \(A\) is ahead of car \(B\) after two-minute duration.

The time cars travel side by side.

d)

Refer the graph of the velocity functions.

When car \(A\) goes faster, the area between the curves \(A\) and \(B\) for the time interval, \(t = 0\) to \(t = 1\) min shows the distance the car \(A\), which is ahead by 3 squares.

Let the cars be side by side at time, \(t = x\).

When car \(B\) goes faster, the area between the curves for the time interval, \(t = 0\) to \(t = x\) is same as the area for the time interval, \(t = 0\) to \(t = 1\;{\rm{min}}\).

The cars appear to be side by side at the time of \(x \approx 2.2\;{\rm{min}}\).

Therefore, at \(2.2\;{\rm{min}}\), the cars travel side by side.

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