Question

A car starts at 36 km/hr towards east in 10 min. it itunes towards north and travels at a rate of 18km/hr for next 10 min . what is it's average speed and velocity ?
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Answer 1

Answer :-

Given :-

▪️Car travel for 10 min in East direction with the speed of 36 km / hr

▪️Car travel for 10 in North direction with the speed of 18 km / hr

♦Now :-

◾Distance covered in East direction :-

$= 36 \times \dfrac{ 10}{60}$

$= 36 \times \dfrac{1}{6}$

$= 6 \: km$

◾Distance covered in North direction :-

$= 18 \times \dfrac{10}{60}$

$= 18 \times \dfrac{ 1}{6}$

$= 3 \: km$

◾Average speed :-

$\dfrac{ Total \: Distance}{Total\: Time}$

$= \dfrac{6 + 3}{\frac{20}{60}}$

$=\dfrac{9}{\frac{1}{3}}$

$= 27 \: km/hr$

◾Now let us consider this picture :-

$\begin{picture}(20,30)\put(60,30){\vector(1,0){180}}\put(240,30){\vector(0,1){90}}\put(60,30){\vector(2,1){180}}\put(150,95){ R }\put(245,70){ N }\put(120,20){ E }\end{picture}$

◾Here

▪️N = Distance covered in North.

▪️E = Distance covered in East

▪️R = Displacement from Intial to final point

◾From Pythagorean theorem

R² = E² + N²

$\implies (R)^2 = (6)^2 + (3)^2$

$\implies (R)^2 = 36 + 9$

$\implies (R)^2 = 45$

$\implies R = \sqrt{45}$

$\implies R = 3\sqrt{5}$

◾Now average Velocity

$= \dfrac{Total\: Displacement}{Total\: Time}$

$= \dfrac{ 3\sqrt{5}}{\frac{20}{60}}$

$= \dfrac{3\sqrt{5}}{\frac{1}{3}}$

$= 9\sqrt{5}\:km/hr$

Answer 2

Answer:

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