A car moves with a speed of $$60\,km/hr$$ from point $$A$$ to point $$B$$ and then with the speed of $$40\,km/hr$$ from point $$B$$ to point $$C.$$ Further it moves to a point $$D$$ with a speed equal to its average speed between $$A$$ and $$C.$$ Points $$A,B,C$$ and $$D$$ are collinear and equidistant. The average speed of the car between $$A$$ and $$D$$ is
A.
$$30\,km/hr$$
B.
$$50\,km/hr$$
C.
$$48\,km/hr$$
D.
$$60\,km/hr$$
Answer :
$$48\,km/hr$$
Solution :
Let the points $$A, B, C$$ and $$D$$ be separated by $$1\,km.$$ Then
$$\eqalign{
& {t_{AB}} = \frac{1}{{60}}hr,{t_{BC}} = \frac{1}{{40}}hr \cr
& \therefore < {v_{AC}} > = \frac{{1 + 1}}{{\frac{1}{{60}} + \frac{1}{{40}}}} = 48\,km/hr \Rightarrow {t_{CD}}\frac{1}{{48}}hr. \cr
& {\text{Now}}\, < {v_{AD}} > = \frac{{1 + 1 + 1}}{{\frac{1}{{60}} + \frac{1}{{40}} + \frac{1}{{48}}}} = 48\,km/hr \cr} $$
Releted MCQ Question on Basic Physics >> Kinematics
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