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4 years ago

7

To arrive at its destination on time the bus should have maintained a speed of vv kilometers per hour throughout the journey. In

stead, after going the first third of the distance at vv kilometers per hour, the bus increased its speed and went the rest of the distance at 1.2v1.2v kilometers per hour. As a result, the bus arrived at its destination xx minutes earlier than planned. What was the actual duration of the trip?

1 answer:

yulyashka [42]4 years ago

7 0

Answer:

The answer to the question is

The actual duration of the trip is 8/9 times the normal duration or the bus arrived 1/9 times earlier than it would have at a constant speed of v kph

Step-by-step explanation:

Speed of bus in the first third distance = v kph

Speed of bus for the remaining two thirds = 1.2 v kph

To solve the question, we note that speed = distance/time

therefore time = distance/speed

Let the distance of the destination be D, then

Normal duration at speed v = D/v

Actual duration (D/3)÷v + (2D/3)÷(1.2v) ⇒ (1.2*D/3 + 2D/3)1.2v

= $\frac{D}{3v} +\frac{5D}{9v}$ = (8/9)×D/v

Therefore the actual duration is 8/9 times the normal duration

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On your paper, construct a rectangle on a coordinate plane that satisfies these criteria.

german

Answer:

$A = (5,3)$

$B = (-5,3)$

$C = (-5,-8)$

$D = (5,-8)$

Step-by-step explanation:

Required

$Perimeter = 42$

Construct a rectangle whose perimeter is 42 units and satisfies the given conditions.

First, name the rectangle ABCD.

Such that:

$A = (x_1,y_1)$

$B = (x_2,y_2)$

$C = (x_3,y_3)$

$D = (x_4,y_4)$

For the rectangle to be either horizontal or vertical, then:

$y_1 = y_2$ and $y_3 = y_4$

We have that:

$Perimeter = 42$

Replace perimeter with its formula

$2(AB + BC) = 42$

Divide both sides by 2

$AB + BC = 21$

This implies that, the distance between adjacent sides (through the edges) must be equal to 21

Having said that: a set of coordinates that satisfy the given conditions are:

$A = (5,3)$ -- First quadrant

$B = (-5,3)$ -- Second quadrant

$C = (-5,-8)$ -- Third quadrant

$D = (5,-8)$ -- Fourth quadrant

The above quadrants satisfy the condition:

$y_1 = y_2$ and $y_3 = y_4$

<u>HOW TO KNOW THE PERIMETER IS 42</u>

To do this, we simply calculate the distance between the edges and add them up

<u>Distance is calculated as:</u>

<u></u> $D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$ <u></u>

<u></u>

<u>For AB</u>

$A = (5,3)$

$B = (-5,3)$

$D_1 = \sqrt{(5 - (-5))^2 + (3 - 3)^2}= \sqrt{(10)^2 + (0)^2} = \sqrt{100} = 10$

<u>For BC</u>

$B = (-5,3)$

$C = (-5,-8)$

$D_2 = \sqrt{(-5 - (-5))^2 + (3 - (-8))^2}= \sqrt{(0)^2 + (11)^2} = \sqrt{121} = 11$

<u>For CD</u>

$C = (-5,-8)$

$D = (5,-8)$

$D_3 = \sqrt{(-5 -5)^2 + (-8 - (-8))^2}= \sqrt{(-10)^2 + (0)^2} = \sqrt{100} = 10$

<u>For DA</u>

$D = (5,-8)$

$A = (5,3)$

$D_4 = \sqrt{(5 -5)^2 + (-8 -3)^2}= \sqrt{(0)^2 + (11)^2} = \sqrt{121} = 11$

So, the perimeter is:

$P = D_1 + D_2 + D_3 + D_4$

$P = 10 + 11 + 10 +11$

$P = 42$

See attachment for rectangle

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