MCQ
A person travelling on a straight line moves with a uniform velocity v1 for a distance x and with a uniform velocity v2 for the next equal distance. The average velocity v is given by
Options
\[v = \frac{v_1 + v_2}{2}\]
\[v = \sqrt{v_1 v_2}\]
\[\frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}\]
\[\frac{1}{v} = \frac{1}{v_1} + \frac{1}{v_2}\]
Advertisement Remove all ads
Solution
\[\frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}\]
Velocity is uniform in both cases; that is, acceleration is zero.
\[x = v_1 t_1 \Rightarrow t_1 = \frac{x}{v_1}\]
\[x = v_2 t_2 \Rightarrow t_2 = \frac{x}{v_2}\]
Total displacement, \[x' = 2x\]
Total time, \[t = t_1 + t_2\]
∴ Average velocity, \[v = \frac{x'}{t} = \frac{2 v_1 v_2}{v_1 + v_2}\]
\[\Rightarrow \frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}\]
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads
