A certain race is made up of three stretches: A, B and C, each 2 km long, and to be covered by a certain mode of transport. The following table gives these modes of transport for the stretches, and the minimum and maximum possible speeds (in km/hr) over these stretches. The speed over a particular stretch is assumed to be constant. The previous record for the race is 10 min.
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Mr Tortoise completes the race at an average speed of 20 km/hr. His average speed for the first two stretches is four times that for the last stretch. Find the speed over stretch C.
Average speed to cover complete race = 20 km/hr = $$\frac{2+2+2}{t_1 + t_2 + t_3}$$ (Where $$t_1$$ is time taken to cover the distance A, $$t_2$$ is time taken to cover the distance B and $$t_3$$ is time taken to cover the distance C)
So total time =$$ t_1 + t_2 + t_3 = 18$$ min.
Avg. speed for first two streches = $$\frac{4}{t_1 + t_2}$$
Avg. speed for last strech = $$\frac{2}{t_3}$$
Given: $$\frac{4}{t_1 + t_2} = 4 \times \frac{2}{t_3}$$
Or $$2t_1 + 2t_2 = t_3$$
Or $$t_3 = 12$$ min.
So $$V_3 = \frac{2}{12} \times 60$$ = 10 km/hr
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Praveen Vm
3Â years, 1Â month ago
Let y be the distance of the stretch.
For all 3 laps it's the same distance y.
Now , average speed 20 =(y+y+y / t1+t2+t3)
t1,t2,t3 are the time taken in each lap.
Let x be the speed for the last stretch.
1st stretch y=4x(t1)
2nd y=4x(t2)
3rd y=xt3
Now find t1,t2 and t3 from the above eqns. And substitute it inbthe first one.
We get x=10.
Nithin Thattil
3Â years, 7Â months ago
This answer is difficult to understand
Praveen Vm
3Â years, 1Â month ago
🤣Yeah bro