Question:
A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed v. The average speed is 48 km/h. The value of v is
A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed v. The average speed is 48 km/h. The value of v is
Updated On: Oct 25, 2024
- 56 km/h
- 60 km/h
- 50 km/h
- 48 km/h
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The Correct Option is B
Solution and Explanation
Total distance travelled = 200 m
Total time taken $=\frac{100}{40}+\frac{100}{v}$
Average speed $=\frac{total\, distance\, travelled}{total\, time\, taken}$
$48=\frac{200}{\bigg(\frac{100}{40}+\frac{100}{v}\bigg)}$ or $48=\frac{2}{\bigg(\frac{1}{40}+\frac{1}{v}\bigg)}$
or $\, \, \frac{1}{40}+\frac{1}{v}=\frac{1}{24}$
or $\, \, \frac{1}{v}=\frac{1}{24}-\frac{1}{40}=\frac{5-3}{120}=\frac{1}{60}$
or $ \, \, v = 60\, km/hr$
Total time taken $=\frac{100}{40}+\frac{100}{v}$
Average speed $=\frac{total\, distance\, travelled}{total\, time\, taken}$
$48=\frac{200}{\bigg(\frac{100}{40}+\frac{100}{v}\bigg)}$ or $48=\frac{2}{\bigg(\frac{1}{40}+\frac{1}{v}\bigg)}$
or $\, \, \frac{1}{40}+\frac{1}{v}=\frac{1}{24}$
or $\, \, \frac{1}{v}=\frac{1}{24}-\frac{1}{40}=\frac{5-3}{120}=\frac{1}{60}$
or $ \, \, v = 60\, km/hr$
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The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion.
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