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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Physics
Chapter
Motion in One Dimension

BITSAT 2023 · Paper 1 · Physics · Motion in One Dimension

An object moves with speed v_1, v_2 and v_3 along a line segment A...

An object moves with speed v1,v2v_1, v_2 and v3v_3 along a line segment AB,BCA B, B C and CDC D respectively as shown in figure. Where AB=BCA B=B C and AD=3ABA D=3 A B, then average speed of the object will be:
An object moves with speed v_1, v_2 and v_3 along a line segment A B, B C and C D respectively as shown in figure.

Options

  1. A\frac{v_1 v_2 v_3}{3\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}
  2. B\frac{3 v_1 v_2 v_3}{\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}
  3. C\frac{\left(u_1+z_2+v_3\right)}{3}
  4. D(v1+v2+v3)3v1v2v3\frac{\left(v_1+v_2+v_3\right)}{3 v_1 v_2 v_3}

Correct answer

B. \frac{3 v_1 v_2 v_3}{\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}

Step-by-step solution

Consider,AB=xBC=x2x+CD=3x⇒CD=3x−2x=x\mathrm{AB}=x \mathrm{BC}=x 2 x+\mathrm{CD}=3 x \Rightarrow \mathrm{CD}=3 x-2 x=xAverage speed of the object <v>= Total distance  Total time <v>=3xxv1+xv2+xv3=3v1v2v3v2v3+v1v3+v1v2< \mathrm{v} > =\frac{\text{ Total distance }}{\text{ Total time }} < \mathrm{v}>=\frac{3 x}{\frac{x}{v_1}+\frac{x}{v_2}+\frac{x}{v_3}}=\frac{3 v_1 v_2 v_3}{v_2 v_3+v_1 v_3+v_1 v_2}
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