1. ## Distance & Speed

A couple of questions to be solved...

1. Bharat sees Jovie. He estimates that she is 25 leaps away. Jovie sees Bharat and starts running away from him to avoid the meeting. If in every minute, Bharat makes 5 leaps and Jovie makes 6 leaps And 1 leap of Bharat is equal to 2 leaps of Jovie. Find the time in which Bharat meets Jovie (assume an open field with no trees) ?

ans: 12.5 minutes

2. A boat went down the river for a distance of 20 km. It then turned back and returned to its starting point, having travelled a total of 7 hours. On its return trip, at a distance of 12 km from the starting point, it encountered a log, which had passed the starting point at the moment at which the boat had started downstream. The downstream speed of the boat is...?

ans: 10 km/hr

2. Hello, MathLearner!

Here's #2 . . . a challenging problem!

2. A boat went down the river for a distance of 20 km.
It then turned back and returned to its starting point, having travelled a total of 7 hours.
On its return trip, at a distance of 12 km from the starting point, it encountered a log,
which had passed the starting point at the moment at which the boat had started downstream.
Find the downstream speed of the boat.

Let $b$ = speed of the boat in still water.
Let $c$ = speed of the current.

The boat went downstream for 20 km at a speed of $(b+c)$ km/hr.
. . This took: $\frac{20}{b+c}$ hours.

The boat went upstrream for 20 km at a speed of $(b-c)$ km/hr.
. . This took: $\frac{20}{b-c}$ hours.
The total trip took 7 hours: . $\frac{20}{b+c} + \frac{20}{b-c} \:=\:7 \quad\Rightarrow\quad 7b^2 - 40b - 7c^2 \:=\:0 \;\;{\color{red}(a)}$

Consider when the boat met the log . . .

The boat had gone downstream 20 km at $(b+c)$ km/hr.
. . This took: $\frac{20}{b+c}$ hours.
Then the boat went upstream 8 km at $(b-c)$ km/hr.
. . This took: $\frac{8}{b-c}$ hours.
Total time: . $\frac{20}{b+c} + \frac{8}{b-c}\text{ hours.}\;\;{\color{blue}[1]}$

During this time, the log traveled 12 km at $c$ km/hr.
. . This took: . $\frac{12}{c}\text{ hours.}\;\;{\color{blue}[2]}$
Equate [1] and [2]: . $\frac{20}{b+c} + \frac{8}{b-c} \:=\:\frac{12}{c}\quad\Rightarrow\quad 12b^2 - 28bc \:=\:0$

Then: . $4b(3b-7c) \:=\:0 \quad\Rightarrow\quad c \:=\:\frac{3}{7}b\;\;{\color{blue}[3]} \quad\hdots\quad \text{since }b \neq 0$

Substitute into (a): . $7b^2 - 40b - 7\left(\frac{9}{49}b^2\right) \:=\:0 \quad\Rightarrow\quad 40b^2 - 280b \:=\:0$

. . $40b(b - 7) \:=\:0 \quad\Rightarrow\quad\boxed{b \:=\:7\text{ km/hr}}$

Substitute into [3]: . $c \:=\:\frac{3}{7}(3) \quad\Rightarrow\quad\boxed{c\:=\:3\text{ km/hr}}$

Therefore, the downstream speed is: . $b + c \:=\:7 + 3 \:=\:{\color{blue}\boxed{10\text{ km/hr}}}$

3. good job Soroban! what an interesting problem.

4. Originally Posted by Soroban
Hello, MathLearner!

Here's #2 . . . a challenging problem!

Let $b$ = speed of the boat in still water.
Let $c$ = speed of the current.

The boat went downstream for 20 km at a speed of $(b+c)$ km/hr.
. . This took: $\frac{20}{b+c}$ hours.

The boat went upstrream for 20 km at a speed of $(b-c)$ km/hr.
. . This took: $\frac{20}{b-c}$ hours.
The total trip took 7 hours: . $\frac{20}{b+c} + \frac{20}{b-c} \:=\:7 \quad\Rightarrow\quad 7b^2 - 40b - 7c^2 \:=\:0 \;\;{\color{red}(a)}$

Consider when the boat met the log . . .

The boat had gone downstream 20 km at $(b+c)$ km/hr.
. . This took: $\frac{20}{b+c}$ hours.
Then the boat went upstream 8 km at $(b-c)$ km/hr.
. . This took: $\frac{8}{b-c}$ hours.
Total time: . $\frac{20}{b+c} + \frac{8}{b-c}\text{ hours.}\;\;{\color{blue}[1]}$

During this time, the log traveled 12 km at $c$ km/hr.
. . This took: . $\frac{12}{c}\text{ hours.}\;\;{\color{blue}[2]}$
Equate [1] and [2]: . $\frac{20}{b+c} + \frac{8}{b-c} \:=\:\frac{12}{c}\quad\Rightarrow\quad 12b^2 - 28bc \:=\:0$

Then: . $4b(3b-7c) \:=\:0 \quad\Rightarrow\quad c \:=\:\frac{3}{7}b\;\;{\color{blue}[3]} \quad\hdots\quad \text{since }b \neq 0$

Substitute into (a): . $7b^2 - 40b - 7\left(\frac{9}{49}b^2\right) \:=\:0 \quad\Rightarrow\quad 40b^2 - 280b \:=\:0$

. . $40b(b - 7) \:=\:0 \quad\Rightarrow\quad\boxed{b \:=\:7\text{ km/hr}}$

Substitute into [3]: . $c \:=\:\frac{3}{7}(3) \quad\Rightarrow\quad\boxed{c\:=\:3\text{ km/hr}}$

Therefore, the downstream speed is: . $b + c \:=\:7 + 3 \:=\:{\color{blue}\boxed{10\text{ km/hr}}}$
Superb Sorobon !!!!

Awesome answer .. Could you help me solve the other problems i ve posted the same way ...

Thanks anyways !!

5. Hello again, MathLearner!

1. Bharat sees Jovie and estimates that she is 25 leaps away.
Jovie sees Bharat and starts running away from him to avoid the meeting.
If in every minute, Bharat makes 5 leaps and Jovie makes 6 leaps,
and 1 leap of Bharat is equal to 2 leaps of Jovie,
Find the time in which Bharat meets Jovie.

"Bharat sees Jovie and estimates that she is 25 leaps away."
. . This doesn't specify whose leaps are measured here: Bharat's or Jovie's.
I will assume these are Bharat's leaps, which are twice Jovie's.
. . Moreover, I'll measure all distances in Jovie's leaps (called Leaps).

At the beginning, they are 50 Leaps apart.

Bharat makes 5 leaps per minute; Jovie makes 6 Leaps per minute.
Since Bharat's are twice Jovie's leaps, he makes 10 Leaps per minute.

Every minute, Bharat will gain $10-6 \:=\:4$ Leaps.

To gain 50 Leaps, it will take: . $\frac{50}{4} \:=\:\boxed{12.5\text{ minutes}}$

6. Yea got it Sorobon !!!

you ve kind of made me confident in distances.. Now its time to move on to another topic... I ll post more..