A car travels first 30 km with a uniform speed of $60 k m h^{- 1}$ and then next 30 km with a uniform speed of $40 k m h^{- 1}$ . Calculate the average speed of the car.

75 k m h^{- 1}

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50 k m h^{- 1}

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48 k m h^{- 1}

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96 k m h^{- 1}

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Solution

The correct option is B $48 k m h^{- 1}$
Total distance covered by car = 60 km
And, total time taken in journey $t = t_{1} + t_{2}$
$t = \frac{s_{1}}{v_{1}} + \frac{s_{2}}{v_{2}}$

$= \frac{30}{60} + \frac{30}{40}$

$= \frac{5}{4} h$
Therefore, Average Speed $= \frac{Total Distance}{Total Time} = \frac{60 k m}{\frac{5}{4} h} = 48 k m / h$

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A car travels first 30 km with a uniform speed of 60 kmh−1 and then next 30 km with a uniform speed of 40 kmh−1. Calculate the average speed of the car.

The correct option is B 48 kmh−1Total distance covered by car = 60 kmAnd, total time taken in journey t=t1+t2t=s1v1+s2v2 =3060+3040 =54 hTherefore, Average Speed=Total DistanceTotal Time=60km54 h=48 km/h

A car travels first 30 km with a uniform speed of $60 k m h^{- 1}$ and then next 30 km with a uniform speed of $40 k m h^{- 1}$ . Calculate the average speed of the car.