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Q. A car covers $\frac{1}{3}$ part of total distance with a speed of $20 \,km \,hr^{-1}$ and second $\frac{1}{3}$ part with a speed of $30 \,km\, hr^{-1}$ and the last $\frac{1}{3}$ part with a speed of $60 \,km \,hr^{-1}$. The average speed of the car is

Motion in a Straight Line

Solution:

$t_1 = \frac{s/3}{20} = \frac{5}{60} ; t_2 = \frac{s/3}{30} = \frac{5}{90}$
$t_3 = \frac{s/3}{60} = \frac{5}{180}$
$v_{av} = \frac{s}{t_1 + t_2 + t_3} $
$= \frac{s}{\frac{s}{60} + \frac{s}{90} + \frac{s}{180}} = 30\, km/h$