You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Differentiation Using Substitution

A point moves...

Question

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.

Open in App

Solution

$P o s i t i o n a l o n g x - a x i s i s g i v e n b y : x = 4 \sin 6 t P o s i t i o n a l o n g y - a x i s i s g i v e n b y : y = 4 (1 - \cos 6 t) a t t = 5 s, x = 4 \sin (6 \times 5) = 4 \sin 30 = 4 \times \frac{1}{2} = 2 m y = 4 (1 - \cos (6 \times 5)) = 4 (1 - \frac{\sqrt{3}}{2}) = 4 \times 0.134 = 0.536 m N e t d i s \tan c e t r a v e l l e d : d = \sqrt{x^{2} + y^{2}} d = \sqrt{4 + 0.287} d = 2.07 m$

Suggest Corrections

Similar questions

Q. A point moves in

x - y

plane according to the law

x = 4 sin 6 t

and

y = 4 (1 - cos 6 t)

. The distance traversed by the particle in

4

seconds is :

(

x

and

y

are in meters)

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. A point moves on the

x - y

plane according to the law

x = a sin ω t

and

y = a (1 - cos ω t)

where

a

and

ω

are positive constants and

t

is in seconds. Find the distance covered in time

t_{0}

Q. A particle moves in x-y plane according to the equations

x = 4 t^{2} + 5 t + 16

and

y = 5 t

where

x, y

are in metre and

t

is in second. The acceleration of the particle is:

Q. A point moves in the plane xy according to the law

x = a sin ω t

y = b cos ω t

. The particle has a trajectory

Join BYJU'S Learning Program

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.

Position along x -axis is given by:x = 4 sin6tPosition along y- axis is given by:y = 4(1-cos6t)at t = 5 s ,x = 4 sin(6×5) = 4 sin 30 = 4×12 =2 my = 4(1-cos(6×5)) = 4(1-32) = 4×0.134 =0.536 mNet distance travelled :d = x2+y2d =4+0.287d = 2.07 m