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3-2.Motion in Plane
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A cyclist starts from centre 0 of a circular park of radius $1\, km$ and, moves along the path $OPRQO$ as shown in figure.
If he maintains constant speed of $10\, ms^{-1}$, what is his acceleration at point $R$ in magnitude and direction ?
Option A
Option B
Option C
Option D
Solution
Here, object performs circular motion. Hence, its acceleration is called centripetal acceleration.
Hence, acceleration at $\mathrm{R}$ is,
$a=\frac{v^{2}}{r}=\frac{(10)^{2}}{1 \mathrm{~km}}=\frac{100}{10^{3}}$
$=0.1 \mathrm{~m} / \mathrm{s}^{2}$ along $\mathrm{RO}$ (towards centre)
Standard 11
Physics
Similar Questions
A particle is rotating in a circle of radius $1\,m$ with constant speed $4\,m / s$. In time $1\,s$, match the following (in $SI$ units) columns.
| Colum $I$ | Colum $II$ |
| $(A)$ Displacement | $(p)$ $8 \sin 2$ |
| $(B)$ Distance | $(q)$ $4$ |
| $(C)$ Average velocity | $(r)$ $2 \sin 2$ |
| $(D)$ Average acceleration | $(s)$ $4 \sin 2$ |
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