On a 100 km road, a car travels the first 50 km at a uniform speed of $30 k m h^{- 1}$ . Assuming constant velocity in second half also, how fast must the car travel for the next 50 km so as to have an average speed of $45 k m h^{- 1}$ for the entire journey?

90 km/h

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60 km/h

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120 km/h

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45 km/h

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Solution

The correct option is A 90 km/h
Total time for the journey, $t = 100 / 45 = 20 / 9 h r s$
Time for first 50 km, $t_{1} = 50 / 30 = 5 / 3 h r$
So, time for remaining journey, $t_{2} = 20 / 9 - 5 / 3 = 5 / 9 h r s$
Speed for remaining journey $v_{2} = \frac{50}{5 / 9} = 90 k m / h$

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On a 100 km road, a car travels the first 50 km at a uniform speed of 30 km h−1. Assuming constant velocity in second half also, how fast must the car travel for the next 50 km so as to have an average speed of 45 kmh−1 for the entire journey?

The correct option is A 90 km/hTotal time for the journey, t=100/45=20/9 hrsTime for first 50 km, t1=50/30=5/3 hrSo, time for remaining journey, t2=20/9−5/3=5/9 hrsSpeed for remaining journey v2=505/9=90 km/h

On a 100 km road, a car travels the first 50 km at a uniform speed of $30 k m h^{- 1}$ . Assuming constant velocity in second half also, how fast must the car travel for the next 50 km so as to have an average speed of $45 k m h^{- 1}$ for the entire journey?

The correct option is A 90 km/h
Total time for the journey, $t = 100 / 45 = 20 / 9 h r s$
Time for first 50 km, $t_{1} = 50 / 30 = 5 / 3 h r$
So, time for remaining journey, $t_{2} = 20 / 9 - 5 / 3 = 5 / 9 h r s$
Speed for remaining journey $v_{2} = \frac{50}{5 / 9} = 90 k m / h$