   Chapter 4.7, Problem 51PS

Chapter
Section
Textbook Problem

Set up an equation and solve each of the following problems (Objective 3).Plane A can travel 1400  miles in 1 hour less than it takes plane B to travel 2000   miles .The rate of plane B is 50   miles per hour greater than the rate of plane A. Find the times and rates of both planes.

To determine

To Find:

The rates and times for both planes.

Explanation

Approach:

For the uniform motion the rate or speed can be expressed as.

s=dt

Here, the travelled distance is d and rate is s and total travel time is t.

Calculation:

Consider the time taken in travelling for plane B is t, then the time taken in travelling for plane A is t1.

The rate of travel for plane A can be expressed as.

sa=1400milest1................(1)

The rate of travel for plane B can be expressed as.

sb=2000milest...................(2)

As mentioned in the problem

sb=50+sa....................(3).

Substitute the value of sbandsa in equation (3).

2000t=50+1400(t1)=50(t1)+1400(t1)2000(t1)=50t(t1)+1400t2000t2000=50t250t+1400t

Further solving.

50t2650t+2000=0t213t+40=0t28t5t+40=0t(t8)5(t8)=0

Further solving.

(t8)(t5)=0

Therefore,

(t8)=0t=8

And,

(t5)=0t=5

Here, both time values can be possible

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