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Question

Physics Question on Motion in a straight line

A car covers the first one-third of a distance x at a speed of 10kmh110\, km\, h^{-1}, the second one-third at a speed of 20kmh120\, km\, h^{-1} and the last one-third at a speed of 60kmh160\, km \,h^{-1}. Find the average speed of the car over the entire distance x.

A

10kmh110\, km\, h^{-1}

B

12kmh112\, km\, h^{-1}

C

18kmh118\, km\, h^{-1}

D

20kmh120\, km\, h^{-1}

Answer

18kmh118\, km\, h^{-1}

Explanation

Solution

For first one-third of distance Distance covered =x3km= \frac{x}{3} \,km Speed =10kmh1.= 10\,km\, h^{-1}. The time taken for the journey, t1=x/310h=x30ht_{1} = \frac{x/3}{10} h = \frac{x}{30} h For the next one-third of distance : Distance covered =x3km= \frac{x}{3} km Speed =20kmh1.= 20\, km\, h^{-1}. The time taken for travel is t2=x/320h=x60ht_{2} = \frac{x/3}{20} h = \frac{x}{60} h For the last one-third of distance : Distance covered =x3km= \frac{x}{3} \,km Speed =60kmh1.= 60 \,km\, h^{-1}. The time taken for travel is t3=x/360h=x180ht_{3} = \frac{x/3}{60} h = \frac{x}{180} h \therefore\quad Average Speed =totaldistancetotaltime=xx30+x60+x180= \frac{total \,distance}{total\, time} = \frac{x}{\frac{x}{30}+\frac{x}{60}+\frac{x}{180}} =180x10x=18kmh1= \frac{180x}{10x} = 18\, km\, h^{-1}