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]

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Solution

$D e a r s t u d e n t, v_{x} = \frac{d x}{d t} = 24 \cos 6 t v_{y} = \frac{d y}{d t} = 24 \sin 6 t v = \sqrt{v^{2} x + v^{2_{}} y} = 24 m / s^{2} d i s \tan c e t r a v e l l e d = v \times t i m e = 24 \times 5 = 120 m R e g a r d s$

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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]

Dear student,vx=dxdt=24cos6tvy=dydt=24sin6tv=v2x+v2y=24m/s2distance travelled=v×time=24×5=120mRegards

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]

$D e a r s t u d e n t, v_{x} = \frac{d x}{d t} = 24 \cos 6 t v_{y} = \frac{d y}{d t} = 24 \sin 6 t v = \sqrt{v^{2} x + v^{2_{}} y} = 24 m / s^{2} d i s \tan c e t r a v e l l e d = v \times t i m e = 24 \times 5 = 120 m R e g a r d s$