The radius vector of a point A relative to the origin varies as r-at- -b t 2 - where a and b are positive constants- Find the equation of trajectory-

The correct option is A y= b a ^ 2 x ^ 2 We have, x = at, y = bt #160;^2 #160; Thus, t = x #160;a #160; Put this value of ...

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

Byju's Answer

Standard XII

Mathematics

Rate of Change

The radius ve...

Question

The radius vector of a point $A$ relative to the origin varies as $\to r = a t^i + b t^{2}^j$ where $a$ and $b$ are positive constants. Find the equation of trajectory.

y = \frac{b}{a^{2}} x^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y = \frac{a^{2}}{b} x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y = \frac{a^{2}}{b} x^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $y = \frac{b}{a^{2}} x^{2}$
We have,
$x = a t, y = b t^{2}$
Thus, $t = \frac{x}{a}$
Put this value of $t$ in $y$ which gives,
$y = b . {(\frac{x}{a})}^{2}$
or $y = \frac{b}{a^{2}} x^{2}$

Suggest Corrections

Similar questions

Q. The radius vector of a point A relative to the origin varies as

\to r = a t^i + b t^{2}^j

where a and b are positive constants. Find the equation of trajectory.

Q. A radius vector of poin

A

relative to the origin varie with time

t

\to r = a t^i - b t^{2}^j

where

a

and

b

are constants. Find the equation of point's trajectory.

Q. If the position vector of a point

P

relative to the origin varies with time as,

\to r = 2 α t^i - β t^{2}^j

, where

α

and

β

are constants, then the equation for trajectory of the particle is

Q. Path traced by a moving particle in space is called trajectory of the particle. Shape of trajectory is decided by the forces acting on the particle. When a coordinate system is associated with a particle's motion, the curve equation in which the particle moves

[y = f (x)]

is called equation of trajectory. It is just giving us the relation among

x

and

y

coordinates of the particle i.e. the locus of particle. To find equation of trajectory of a particle, find first

x

and

y

co-ordinates of the particle as a function of time and eliminate the time factor. The position vector of car w.r.t. its starting point is given as

\to r = a t^i - b t^{2}^j

where

a

and

b

are positive constants. The locus of a particle is

Q. The trajectory of a projectile in a vertical plane is:

y = a x - b x^{2}

,
where

a

and

b

are constants and

x

and

y

are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is

Join BYJU'S Learning Program

The radius vector of a point A relative to the origin varies as →r=at^i+bt2^j where a and b are positive constants. Find the equation of trajectory.

The correct option is A y=ba2x2We have,x=at,y=bt2Thus, t=xaPut this value of t in y which gives,y=b.(xa)2or y=ba2x2

The radius vector of a point $A$ relative to the origin varies as $\to r = a t^i + b t^{2}^j$ where $a$ and $b$ are positive constants. Find the equation of trajectory.

The correct option is A $y = \frac{b}{a^{2}} x^{2}$
We have,
$x = a t, y = b t^{2}$
Thus, $t = \frac{x}{a}$
Put this value of $t$ in $y$ which gives,
$y = b . {(\frac{x}{a})}^{2}$
or $y = \frac{b}{a^{2}} x^{2}$