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)

96 m

Answer 2

Here, $x = 4sin6t,\, y = 4(1 - cos6t)$ $\upsilon_{x} = \frac{dx}{dt} = \frac{d}{dt}\left(4sin6t\right) = 24cos 6t$ $\upsilon _{y} = \frac{dx}{dt} = \frac{d}{dt}4\left(1-cos6t\right) = 24sin6t$ $\upsilon = \sqrt{\upsilon^{2}_{x} +\upsilon^{2}_{y}} = \sqrt{\left( 24cos 6t\right)^{2}+\left(24sin6t\right)^{2}} = 24\,m\,s^{-1}$ i.e., speed of the particle is constant. Hence, the distance traversed by the particle in 4 s is $s = \upsilon t = \left(24 \times 4\right) m = 96 \,m$