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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Motion in Two Dimensions

BITSAT 2013 · Physics · Motion in Two Dimensions

The velocity and acceleration vectors of a particle undergoing...

The velocity and acceleration vectors of a particle undergoing circular motion are v→=2i^m/s\overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s} and a→=2i^+4j^m/\mathrms2\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \mathrm{m} / \mathrms^{2} respectively at an instant of time. The radius of the circle is -

Options

  1. A1m1 \mathrm{ m}
  2. B2m2 \mathrm{ m}
  3. C3m3 \mathrm{ m}
  4. D4m4 \mathrm{ m}

Correct answer

A. 1m1 \mathrm{ m}

Step-by-step solution

It can be observed that component of acceleration perpendicular to velocity is a=4m/s2∴ radius =v2ac=(2)24=1\mathrm{a}=4 \mathrm{ m} / \mathrm{s}^{2} \therefore \text{ radius }=\frac{\mathrm{v}^{2}}{\mathrm{a}_{\mathrm{c}}}=\frac{(2)^{2}}{4}=1 metre \text{ metre }
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