Question

A boat leaves for its destination to B) 60 kms away traveling at a speed of 15 km/hr at 800 am
From the opposite skie, another boat traveling at 24 kmhr approaches. The other boat had started at
100 am tom point Al what distance from point is would they meet (Consider that the water is still?​

Answer 1

Answer:

sorry i don't know its answer

Answer 2

Answer:

x = 18.46

Step-by-step explanation:

Given, the boat(1) is travelling to point (B) which is 60 kms from initial point with speed ($v_1$) = 15 kmph, and it starts at 8:00 am.

And another boat(2) is coming from point (B) towards boat(1) with speed($v_2$) = 24 kmph, which had started at 10:00 am.

Now let the both boats meet at a distance "x" from the point (B)

For this at first lets calculate how much distance does the boat(1) travel before the clock ticks 10:00 am ,

As the clock to tick 10:00 am from 8:00 am it takes 2 hrs.

So, t = 2 hrs and speed ($v_1$) = 15 kmph

Hence, distance(d) = speed($v_1$) × time(t)

          ⇒ d = 15 kmph × 2hrs

          ⇒ d = 30 kms(lets call this point as some C)

Now, if the boat(2) has travelled "x" distance from point (B)

Then, the boat(1) has travelled ((60-30) - x) i.e. (30 - x)

For, boat(1):

30 - x = $v_1$ × t

⇒ $\frac{30 -x}{15} = t$......(1)

For boat(2)

x = $v_2$ × t (here the time remains same because the journey of the two boats start from 10:00 am and at "t" time the journey completes according to us).

⇒ x = 24 t....(2)

substituting equation (1) in (2) we get:

$x = 24\times\frac{30-x}{15}$

Therefore, x = 18.46 kms from point (B) the two boats say hello!

#SPJ2