Question

A particle moves in xy plane according to the law x = 4sin6t and y = 4(1 - cos6t). The distance traversed by the particle in 4 s is (x and y are in metres)

• A. 96 m
• B. 48 m
• C. 24 m
• D. 108 m

Answer 1

Answer: (A)

Here, $x=4sin6t,y=4(1-cos6t)$
${\upsilon }_x=\frac{dx}{dt}=\frac{d}{dt}\left(4sin6t\right)=24cos6t$
${\upsilon }_y=\frac{dx}{dt}=\frac{d}{dt}4\left(1-cos6t\right)=24sin6t$
$\upsilon =\sqrt{{\upsilon }_x^2+{\upsilon }_y^2}=\sqrt{{\left(24cos6t\right)}^2+{\left(24sin6t\right)}^2}=24m{s}^{-1}$
i.e., speed of the particle is constant. Hence, the distance traversed by the particle in 4 s is
$s=\upsilon t=\left(24\times 4\right)m=96m$