A car travels a distance of ' x ' with speed v₁ and then same distance ' x ' with speed v₂ in the same direction. The average speed of the car is: (a) (v₁ v₂)/(2(v₁ + v₂)) (b) (2v₁ v₂)/(v₁ + v₂) (c) (2x)/(v₁ + v₂) (d) (v₁ + v₂)/(2)

Answer: (b). Total distance 2x, total time x/v₁ + x/v₂ = (x(v₁ + v₂))/(v₁ v₂). vavg = 2x/t = (2v₁ v₂)/(v₁ + v₂) So the answer is (b).

A car travels a distance of ' x ' with speed v₁ and then same… | JEE Main 2023 (25 Jan, Shift 1)

Q12JEE Main 2023 · 25 JAN · SHIFT 1Easy

Distance, Displacement & Average Speed

A car travels a distance of ' $x$ ' with speed $v_1$ and then same distance ' $x$ ' with speed $v_2$ in the same direction. The average speed of the car is:

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Answer: (b)

Total distance $2x$ , total time $\frac{x}{v_1} + \frac{x}{v_2} = \frac{x(v_1 + v_2)}{v_1 v_2}$ .
$v_{avg} = \frac{2x}{t} = \frac{2v_1 v_2}{v_1 + v_2}$
So the answer is (b).

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Distance, Displacement & Average Speed question from JEE Main 2023 (25 Jan, Shift 1)

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