Question

9th physics homework sum-A train travels first 15 km at a uniform speed 30 km/h.next 75 km at uniform speed 50 km/h.last 10 km at a uniform speed of 20 km/h.calculate the average speed.

Answer 1

Distance D1 = 15 km     Speed V1 = 30 kmph      Duration T1 = 15/30 = 0.50 hrs
Distance D2 = 75 km     speed V2 = 50kmph       duration T2 = 75/50 = 1.5 hrs
Distance D3 = 10 km     speed V3 = 20 kmph      duration T3 = 10/20 = 0.5 hrs
 total      D = 100 km                                                      total  =  2 hours

   average speed = V = total distance / total time duration
                                 = 100 km / 2 hrs = 50 kmph

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We can also do this by using the formula:

$V=\frac{total\ distance}{total\ time\ duration}\\\\.\ \ \ \ =\frac{D1+D2+D3}{T1+T2+T3}=\frac{D1+D2+D3}{\frac{D1}{V1}+\frac{D2}{V2}+\frac{D3}{V3}}\\\\V=\frac{15+75+10\ km}{0.50+1.50+0.50\ hrs}=\frac{100\ km}{2\ hrs}=50\ kmph$

If you can use the formula, that is fine. Other wise do that in the first method.

Answer 2

Here is your solution

Given :-

First 15 km at a uniform speed of 30km/h.

Next 75km at at a uniform speed of 50km/h.

Last 10km/h at a uniform speed of 20km/h.

Time 1st = distance /speed = 15/30
Time 2nd = distance /speed = 75/50
Time 3rd = distance /speed = 10/20

$total \: time = \frac{15}{30} + \frac{75}{50} + \frac{10}{20} \\ \\ total \: time= \frac{150 + 450 + 150}{300} \\ \\ total \: time \: = \frac{750}{300} = \frac{5}{2}$
Total distance = 15km+75km+10km = 100km

$average \: speed = \frac{total \: distance}{total \: time} \\ average \: speed = \frac{100}{ \frac{5}{2} } = \frac{100 \times 2}{5}\\average \: speed = 40km/h$

Hence,
Average speed of train is 40km/h

Hope it helps you