A car covers a distance at a speed of 60 km | vedclass

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Question

(A) Let the car cover a distance $s$ from point $A$ to $B$ and then return to point $A$.The average speed is defined as the total distance covered div

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) Let the car cover a distance $s$ from point $A$ to $B$ and then return to point $A$.The average speed is defined as the total distance covered divided by the total time taken.Time taken from $A$ to $B$ is $t_1 = \frac{s}{60}$.Time taken from $B$ to $A$ is $t_2 = \frac{s}{V}$.The total distance is $s + s = 2s$.The total time is $t_1 + t_2 = \frac{s}{60} + \frac{s}{V}$.Given the average speed is $48 \ km \ h^{-1}$,we have:$48 = \frac{2s}{\frac{s}{60} + \frac{s}{V}}$Dividing both sides by $s$ (assuming $s 
eq 0$):$48 = \frac{2}{\frac{1}{60} + \frac{1}{V}}$$\frac{1}{60} + \frac{1}{V} = \frac{2}{48} = \frac{1}{24}$$\frac{1}{V} = \frac{1}{24} - \frac{1}{60}$$\frac{1}{V} = \frac{5 - 2}{120} = \frac{3}{120} = \frac{1}{40}$Therefore,$V = 40 \ km \ h^{-1}$.

$A$ car covers a distance at a speed of $60 \ km \ h^{-1}$. It returns and comes back to the original point moving at a speed of $V$. If the average speed for the round trip is $48 \ km \ h^{-1}$,then the magnitude of $V$ is (in $km \ h^{-1}$)

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