Question

Two motorists met at 10

a.M. At the dadar railway station. After their meeting, one of them proceeded in the east direction while the other proceeded in the north direction. Exactly at noon, they were 60 km apart. Find the speed of the slower motorist if the difference of their speeds is 6 km/h.

Answer 1

Answer:

Speed of slowest motorist is 7.17 km/h

Explanation:

As we know that the two motorist moves perpendicular to each other

They meet at 10 AM and after 4 hours exactly at noon the separation between them is 60 km

so let say the distance moved by each motorist is given as the product of their speed and time

so here we have

$d_1 = v_1(4)$

$d_2 = v_2(4)$

now we know that

$v_1 - v_2 = 6 km/h$

also we know

$\sqrt{16(v_1^2 + v_2^2)} = 60 km$

$v_1^2 + v_2^2 = 225$

$(6 + v_2)^2 + v_2^2 = 225$

$v_2 = 7.17 km/h$

$v_1 = 13.17 km/h$

#Learn

Topic : relative speed

https://brainly.in/question/13865425

Answer 2

Answer:

The Distance can be shown as the hypotenuse of a right angle triangle

Now, we need to find the Pythagoras Triplet suitable for this. And 18, 24, 30 would be the required triplets.

So, Speed of the slower motorist = 18 kmph