A point moves in the plane xy according to the law x-sin- t- y-1-cos- -t- where - and - are positive constants and t is time- The distance traversed by the point in time t0 is 4- t0x- Find x-

Name: CENGA - Grade 11 - Physics - Motion in two Dimension - Q39
Uploaded: 2022-07-04T10:36:42+05:30
Description: Given, x=αsin⁡ω t nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;v_x=dxdt nbsp; nbsp; ...

Given, x=αsin⁡ω t nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;v_x=dxdt nbsp; nbsp; ...

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

Byju's Answer

Standard VII

History

Battles

A point moves...

Question

A point moves in the plane $x y$ according to the law $x = α sin ω t$ , $y = α (1 - cos ω t)$ , where $α$ and $ω$ are positive constants and $t$ is time. The distance traversed by the point in time $t_{0}$ is $\frac{4 α ω t_{0}}{x}$ . Find $x$ .

Open in App

Solution

Given, $x = α sin ω t$
$v_{x} = \frac{d x}{d t}$
$= α cos ω t \cdot ω$
$= α ω cos ω t$

$y = α (1 - cos ω t)$
$v_{y} = \frac{d y}{d t} = α [0 - (- sin ω t) ω] = α ω sin ω t$

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = α ω =$ constant

Distance in time $t_{0}$
$s = v t_{0}$
$= α ω t_{0}$

Suggest Corrections

Similar questions

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 point moves in xy plane, according to law x=kt and y= kt(1-

α

t) where k and

α

are positive constants and t is time. Find the time after which the angle between the velocity and acceleration is

45^{\circ}

Q. A particle moves in the XY-plane according to the law

x = k t, y = k t (1 - α t)

where

k

and

α

are positive constants and

t

is time. The trajectory of the particle is:

Q. A point moves according to the law

x = a t, y = a t (1 - α t)

where

a

and

α

are positive constants and

t

is time. If the moment at which angle between velocity vector and acceleration vector is

\frac{π}{4}

is given as

t = \frac{x}{α}

. Find

x

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)

Join BYJU'S Learning Program

Explore more

Battles

Standard VII History

Join BYJU'S Learning Program

A point moves in the plane xy according to the law x=αsinω t, y=α(1−cosω t), where α and ω are positive constants and t is time. The distance traversed by the point in time t0 is 4αω t0x. Find x.

Given, x=αsinω t vx=dxdt =α cosω t⋅ω =αωcosω t y=α(1−cosω t) vy=dydt=α[0−(−sinω t)ω]=αωsinω t v=√v2x+v2y=αω=constant Distance in time t0 s=vt0 =αω t0

A point moves in the plane $x y$ according to the law $x = α sin ω t$ , $y = α (1 - cos ω t)$ , where $α$ and $ω$ are positive constants and $t$ is time. The distance traversed by the point in time $t_{0}$ is $\frac{4 α ω t_{0}}{x}$ . Find $x$ .