Question

A boat travels 50km east, the 120km north and finally it comes back to the straight point through the shortest distance. The total time of journey is 3h.what is the average speed in kmh-1 , over the entire trip?

Answer 1

Answer:

The Boat travelled the total distance with an average speed of 100 km/h.

Explanation:

We know that,

The boat moves 50 km towards the East, then to North 120 km.

Now the shortest distance between those 2 points would be its displacement which is a straight line,

Now, if we graph this, we would get a Triangle.

(You can refer above)

Let the starting point be A, then it moves East 50km, let it end at B, then from B it moves North 120km to point C.

Now, AC will be a straight line moving South west, as it is the displacement.

Now, we know that, directions are perpendicular to it adjacent directions

For ex:-

North is perpendicular to East and West.

South is perpendicular to East and West.

Thus,

∠B = 90° [East is perpendicular to North]

Hence, ABC is a right angled triangle

Thus,

By Pythagoras theorem

AB² + BC² = AC²

AC² = 50² + 120²

AC² = 2500 + 14400

AC² = 16900

AC = √16900

AC = 130km

Now,

Average Speed = Total Distance/Total Time

Total Distance = 50 + 120 + 130

= 250 km

Total Time = 3 hrs

Hence,

S(av.) = 300/3

S(av.) = 100 km/h

Thus,

The Boat travelled the total distance with an average speed of 100 km/h.

Hope it helped and you understood it........All the best.

Answer 2

★ Solution :-

According to the given question ,

The boat moves 50 km towards the east (AB)

Then, the boat moves 120 km towards the north

(AC )

Then ,the boat comes back to it's starting point through the shortest distance

Here , the boat is moving in a right angled triangle so the shortest distance will be it's hypotenuse

In ∆ ABC ,

$\implies \mathtt{{ CB }^{2} = {AB }^{2} + {AC }^{2} }$

$\implies \mathtt{{ CB }^{2} = {50 }^{2} + {120 }^{2} }$

$\implies \mathtt{{ CB }^{2} = 2500+ 14400 }$

$\implies \mathtt{{ CB }^{2} = 16900 }$

$\implies \mathtt{CB = \sqrt{ 16900 } }$

$\implies \mathtt{CB = 130 \: km}$

_______________

Total distance covered by the boat (d)

$\implies\mathtt{d = AB + AC + CB }$

$\implies\mathtt{d = 50 + 120 + 130 }$

$\implies\mathtt{d = 300 \: km}$

• Total distance covered by the boat =300 km

• Total time taken = 3 h

We need to find the average speed of the boat over the entire trip

we know that ,

$\bigstar\boxed{\mathtt{ Avg .speed = \dfrac{Total \: distance}{total \: time}}}$

Now , let's substitute the given values in the above equation ,

$\implies\mathtt{ Avg .speed = \dfrac{300}{3}}$

$\implies \boxed{\mathtt{ Avg .speed = 100 \dfrac{km}{hr} }}$

The average speed of the boat over the entire trip is 100 km /hr