A train travels at a speed of $50 K m / h r$ for $0.5 h r$ , at $30 K m / h r$ for next $0.26 h r$ and then $70 K m / h r$ for the next $0.76 h r$ . What is the average speed of the train:

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Solution

$v_{1} = 50 k m p h$
$t_{1} = 0.5 h r$
$d_{1} = 50 \times 0.5 = 25 k m$
$v_{1} = 30 k m p h$
$t_{1} = 0.26 h r$
$d_{1} = 30 \times 0.26 = 7.8 k m$
$v_{1} = 70 k m p h$
$t_{1} = 0.76 h r$
$d_{1} = 70 \times 0.76 = 53.2 k m$
Average speed will be,
$V = \frac{d_{1} + d_{2} + d_{3}}{t_{1} + t_{2} + t_{3}}$

$V = \frac{25 + 7.8 + 53.2}{0.5 + 0.26 + 0.76}$

$V = \frac{86}{1.52}$
$V = 56.57 k m p h$

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A train travels at a speed of 50 Km/hr for 0.5 hr, at 30 Km/hr for next 0.26 hr and then 70 Km/hr for the next 0.76 hr. What is the average speed of the train:

v1=50kmpht1=0.5hrd1=50×0.5=25kmv1=30kmpht1=0.26hrd1=30×0.26=7.8kmv1=70kmpht1=0.76hrd1=70×0.76=53.2kmAverage speed will be,V=d1+d2+d3t1+t2+t3V=25+7.8+53.20.5+0.26+0.76V=861.52V=56.57kmph

A train travels at a speed of $50 K m / h r$ for $0.5 h r$ , at $30 K m / h r$ for next $0.26 h r$ and then $70 K m / h r$ for the next $0.76 h r$ . What is the average speed of the train: