A car moves towards north at a speed of 54 ,k | vedclass

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(B) Given: Speed $v = 54 \,km/h = 54 	imes \frac{5}{18} \,m/s = 15 \,m/s$. Time $t_1 = 1 \,h$,$t_2 = 1 \,h$. Total time $T = 2 \,h$.Distance covered

Vedclass · Accepted Answer

$15 \,m/s, \frac{15}{\sqrt{2}} \,m/s$

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(B) Given: Speed $v = 54 \,km/h = 54 	imes \frac{5}{18} \,m/s = 15 \,m/s$. Time $t_1 = 1 \,h$,$t_2 = 1 \,h$. Total time $T = 2 \,h$.Distance covered in north direction $d_1 = v 	imes t_1 = 54 	imes 1 = 54 \,km$.Distance covered in east direction $d_2 = v 	imes t_2 = 54 	imes 1 = 54 \,km$.Total distance $= d_1 + d_2 = 54 + 54 = 108 \,km$.Average speed $= \frac{	ext{Total distance}}{	ext{Total time}} = \frac{108 \,km}{2 \,h} = 54 \,km/h = 15 \,m/s$.Displacement is the vector sum of the two displacements: $\vec{s} = \sqrt{d_1^2 + d_2^2} = \sqrt{54^2 + 54^2} = 54\sqrt{2} \,km$.Average velocity $= \frac{	ext{Total displacement}}{	ext{Total time}} = \frac{54\sqrt{2} \,km}{2 \,h} = 27\sqrt{2} \,km/h$.Converting to $m/s$: $27\sqrt{2} 	imes \frac{5}{18} = \frac{3\sqrt{2} 	imes 5}{2} = \frac{15\sqrt{2}}{2} = \frac{15}{\sqrt{2}} \,m/s$.Thus,the average speed is $15 \,m/s$ and average velocity is $\frac{15}{\sqrt{2}} \,m/s$.

$A$ car moves towards north at a speed of $54 \,km/h$ for $1 \,h$. Then it moves eastward with the same speed for the same duration. The average speed and velocity of the car for the complete journey is ..........

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