Question

Two boat landings are 1.0 km apart on the same bank of a stream that flows at 1.4 km/h. A motorboat makes the round trip between the two landings in 50 minutes. What is the speed of the boat relative to the water?

Answer 1

Here again you have to use the same concept.

Answer 2

Let the speed of the boat wrt the river be x, then actual speed of boat would become (x+1.4) and (x-1.4).

Answer 3

Can you find the value of x now ?

Answer 4

I'm not sure I understand this one

Answer 5

you mean you don't understand ?

Answer 6

yeah

Answer 7

Did u understood upto what i wrote above ?

Answer 8

yes, i understand what you drew, but I'm still confused on how to find the value of x

Answer 9

brb !

Answer 10

\(\color{blue}{\text{Originally Posted by}}\) @Abhisar
Now, total time taken to cover the round trip is 50 mins = 0.83 hours.
\(\color{blue}{\text{End of Quote}}\)

Answer 11

Time = Distance/speed.
So,

\(\huge\sf \frac{1}{(x+1.4)}+\frac{1}{(x-1.4)}=0.83\)
Solve it and find the value of x

Answer 12

Getting it ?

Answer 13

yeah, so far

Answer 14

can you calculate now ?

Answer 15

Remember a^2 - b^2 = (a - b)(a + b)

Answer 16

so would I have to multiply both sides by (x+1.4)(x-1.4) to get rid of the fractions

Answer 17

Then I would get
\[(x+1.4)+(x-1.4)=.83(x+1.4)(x-1.4)\]?

Answer 18

\(\sf (x+1.4)(x-1.4)=x^2-(1.4)^2\)

Answer 19

I used the identity \(\sf a^2-b^2=(a+b)(a-b)\)

Answer 20

but I'm still solving for x, right?

Answer 21

yes. Add the left hand side of the equation@Abhisar



\(\huge\sf \frac{1}{(x+1.4)}+\frac{1}{(x-1.4)}=0.83\)

Answer 22

so the answer I get when I do that problem is \[x ^{2}-1.96\]

Answer 23

It will be the denominator of the left hand side after adding. Find numerator too

Answer 24

\(\huge\sf \frac{(x+1.4)+(x-1.4)}{x ^{2}-1.96}=0.83\)

Now find x

Answer 25

wouldn't it be easier to factor the bottom then find x, or can you not do that?

Answer 26

Do however you like. I find this method easier. The answer to your question will be the value of x

Answer 27

okay so I would multiply both sides by \[x ^{2}-1.96\] to get \[(x ^{2}-1.96)+(x+1.4)+(x-1.4)=0.83(x ^{2}-1.96)\] right?

Answer 28

Then I would add everything up on the left hand side and get \[(x ^{2}+2x-1.96)=.83x ^{2}-1.6268\]

Answer 29

That's where I get stuck

Answer 30

\(\huge\sf \frac{(x+1.4)+(x-1.4)}{x ^{2}-1.96}=0.83\)

--> 2x= \(\sf 0.83x^2-1.62\)

-->\(\sf 0.83x^2-2x-1.62=0\)

Finally you will have to solve the quadratic equation to get the value of x. `This is the easiest way.`

Answer 31

okay, hold on

Answer 32

okay I set it up as \[x=\frac{ 2\pm \sqrt{(-2)^{2}-4(0.83)(-1.62)} }{ 2(.83) }\] teh when I type it all in, my answer is \[x=\frac{ 2\pm \sqrt{9.3784} }{ 1.66 }\]

Answer 33

it will be 3 m/s. But do cross check.

Answer 34

yeah, thats what I got

Answer 35

So, that's it? The speed is 3m/s?

Answer 36

yes i think so.

Answer 37

what if I had the same speed boats but the distance was 6 instead of 1. How would that affect the equation?

Answer 38

Do the same way

Answer 39

\(\color{blue}{\text{Originally Posted by}}\) @Abhisar
yes. Add the left hand side of the equation@Abhisar



\(\huge\sf \frac{1}{(x+1.4)}+\frac{1}{(x-1.4)}=0.83\)
\(\color{blue}{\text{End of Quote}}\)


just put 6 at the place of 1 in this equation.

Answer 40

and it also says the measurement is in km/h, so it's actually not 3m/s it's still 3km/h?

Answer 41

it says that's the right answer. thank you so much!

Answer 42

\(\color{red}{\huge\bigstar}\huge\text{You're Most Welcome! }\color{red}\bigstar\)
\(~~~~~~~~~~~~~~~~~~~~~~~~~~~\color{green}{\huge\ddot\smile}\color{blue}{\huge\ddot\smile}\color{pink}{\huge\ddot\smile}\color{red}{\huge\ddot\smile}\color{yellow}{\huge\ddot\smile}\)