A point moves in x-y plane according to the law x=4 sin 6t and y=4(1-cos 6t). The distance traversed by the particle in 4 seconds is (x and y are in meters)

Open in App

Solution

x=4 sin 6t; y=4(1-cos 6t)
$\frac{d x}{d t} = v_{x}$ =4.6 cos 6t; $\frac{d y}{d t} = v_{y}$ =4.6 sin 6t
Now speed = $| \to v | = \sqrt{v_{x}^{2} + v_{y}^{2}}$
$= \sqrt{(24 c o s 6 t)^{2} + (24 s i n 6 t)^{2}}$
Speed = 24 m/s
Oh wow!
Speed is constant through the journey
So now it's easy
I know speed = $\frac{d i s t a n c e}{t i m e}$
24= $\frac{d i s t a n c e}{4}$
Distance = 96 m

Suggest Corrections

Similar questions

A point moves in x-y plane according to the law x=4 sin 6t and y=4(1-cos 6t). The distance traversed by the particle in 4 seconds is (x and y are in meters)

Q. How to solve
Q.25. A point moves in x-y plane according to the law x = 4 sin 6 t any y = 4 (1 - cos 6 t). Find distance traversed by the particle in 5 seconds, when x and y are in metres.
[Ans. 120 m]

A point moves in x-y plane according to the law x=4 sin 6t any y=4(1-cos6t).Find the distance traversed by the particle in 5 seconds,when x and y are in metres.

A point moves in the xy plane according to the law x=a sin $ω$ t, y=a(1-cos $ω$ t), where a and $ω$ are positive constants.
(a)Find the distance s traversed by the point during the time t;
(b) Find the equation of trajectory.