A car moves a distance of 200 ; m. It covers | vedclass

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(B) Let the total distance be $d = 200 \; m$. The first half distance is $d_1 = 100 \; m$ and the second half distance is $d_2 = 100 \; m$.The time ta

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) Let the total distance be $d = 200 \; m$. The first half distance is $d_1 = 100 \; m$ and the second half distance is $d_2 = 100 \; m$.The time taken for the first half is $t_1 = \frac{d_1}{v_1} = \frac{100}{40} \; h$.The time taken for the second half is $t_2 = \frac{d_2}{v} = \frac{100}{v} \; h$.The average speed $v_{avg}$ is given by the formula: $v_{avg} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{d}{t_1 + t_2}$.Substituting the given values: $48 = \frac{200}{\frac{100}{40} + \frac{100}{v}}$.Dividing the numerator and denominator by $100$: $48 = \frac{2}{\frac{1}{40} + \frac{1}{v}}$.Rearranging the terms: $\frac{1}{40} + \frac{1}{v} = \frac{2}{48} = \frac{1}{24}$.Solving for $v$: $\frac{1}{v} = \frac{1}{24} - \frac{1}{40} = \frac{5 - 3}{120} = \frac{2}{120} = \frac{1}{60}$.Therefore,$v = 60 \; km/h$.

$A$ car moves a distance of $200 \; m$. It covers the first half of the distance at a speed of $40 \; km/h$ and the second half of the distance at a speed of $v$. If the average speed is $48 \; km/h$,the value of $v$ is: (in $; km/h$)

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