Calculus: Graphical, Numerical, Algebraic

3rd Edition

ISBN: 9780133688399

Author: Finney, Ross L.

Publisher: Pearson/Prentice Hall

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Question

Chapter 2.1, Problem 70E

(a)

To determine

To find: The value of g .

(a)

Expert Solution

Answer to Problem 70E

The value of g is 1.25 .

Explanation of Solution

Given information: On a small airless planet a rock is dropped from rest to fall is represented by y=gt2 m in t sec , g a constant. Assume that the rock falls 20 m below to the bottom of a crevasse and reaches the bottom in 4 sec .

Calculation:

As given that rock travels 20 m in 4 sec .

Substitute 20 for y and 4 for t in the given equation.

y=gt220=g×4220=16gg=1.25

Therefore, the value of g is 1.25 .

(b)

To determine

To find: The average speed for the fall.

(b)

Expert Solution

Answer to Problem 70E

The average speed for the fall is 5 m/sec .

Explanation of Solution

Calculation:

Use the formula for the speed of the object.

s=ΔyΔt

Now substitute 4 for t in the equation of path for the distance travelled after 3 seconds.

y=gt2y=1.25t2y=1.25×42y=20

Initially the object was at rest so y=0 at t=0 .

Now substitute these values in formula for speed.

s=ΔyΔt=y2−y1t2−t1=20−04−0=5 m/sec

Therefore, the average speed for the fall is 5 m/sec .

(c)

To determine

To find: The speed of the rock when it hits the bottom.

(c)

Expert Solution

Answer to Problem 70E

The speed of the rock when it hits the bottom is 10 m/sec .

Explanation of Solution

Calculation:

The speed of the rock when it hits the bottom is the instantaneous speed of the rock at t=4 sec .

The instantaneous speed of an object at t=a seconds is the average speed from time t=a seconds to t=a+h seconds.

Use the formula for average speed.

s=ΔyΔt=y2−y1t2−t1

Substitute 4+h for t in the given equation for the distance travelled.

y2=1.25(4+h)2

Substitute 4 for t in the given equation for the distance travelled.

y1=1.25(4)2=1.25×16=20

Substitute these values in the formula for average speed.

s=y2−y1t2−t1=1.25(4+h)2−204+h−4=1.25(4+h)2−20h=1.25(16+h2+8h)−20h

Further simplify,

s=20+1.25h2+10h−20h=h(1.25h+10)h=1.25h+10

Substitute h=0 for the exact instantaneous speed.

s=1.25×0+10=10 m/sec

Therefore, the speed of the rock when it hits the bottom is 10 m/sec .

Chapter 2.1, Problem 69E

Chapter 2.1, Problem 71E

Chapter 2 Solutions

Calculus: Graphical, Numerical, Algebraic

Ch. 2.1 - Prob. 1QR Ch. 2.1 - Prob. 2QR Ch. 2.1 - Prob. 3QR Ch. 2.1 - Prob. 4QR Ch. 2.1 - Prob. 5QR Ch. 2.1 - Prob. 6QR Ch. 2.1 - Prob. 7QR Ch. 2.1 - Prob. 8QR Ch. 2.1 - Prob. 9QR Ch. 2.1 - Prob. 10QR

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - 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