The co-ordinates of a particle moving in plane is given by, The motion of the particle is

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Q.
The co-ordinates of a particle moving in $x - y$ plane is given by $x = 2 + 4 t, y = 3 t + 8 t^{2}$

The motion of the particle is [2024]

1 uniform motion along a straight line

2 uniformly accelerated having motion along a parabolic path

3 non-uniformly accelerated

4 uniformly accelerated having motion along a straight line

Ans.
(2)

Given, $x = 2 + 4 t$ and $y = 3 t + 8 t^{2}$

$v_{x} = \frac{d x}{d t} = 4 =$ constant, hence $a_{x} = 0$

and $v_{y} = 3 + 16 t$ also, $\frac{d v_{y}}{d t} = a_{y} = 16$

the motion will be uniformly accelerated motion.

Also, $t = \frac{x - 2}{4}$

$∴ y = 3 (\frac{x - 2}{4}) + 8 {(\frac{x - 2}{4})}^{2}$

This is a quadratic equation so path will be parabola, parabolic path.

Solution

Q.
The co-ordinates of a particle moving in $x - y$ plane is given by $x = 2 + 4 t, y = 3 t + 8 t^{2}$

The motion of the particle is [2024]

1 uniform motion along a straight line

2 uniformly accelerated having motion along a parabolic path

3 non-uniformly accelerated

4 uniformly accelerated having motion along a straight line

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Solution

Q. The co-ordinates of a particle moving in x-y plane is given by x=2+4t,y=3t+8t2 The motion of the particle is [2024]

1 uniform motion along a straight line

2 uniformly accelerated having motion along a parabolic path

3 non-uniformly accelerated

4 uniformly accelerated having motion along a straight line

Ans. (2) Given, x=2+4t and y=3t+8t2 vx=dxdt=4= constant, hence ax=0 and vy=3+16t also, dvydt=ay=16 the motion will be uniformly accelerated motion. Also, t=x-24 ∴ y=3(x-24)+8(x-24)2 This is a quadratic equation so path will be parabola, parabolic path.

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Q.
The co-ordinates of a particle moving in $x - y$ plane is given by $x = 2 + 4 t, y = 3 t + 8 t^{2}$

The motion of the particle is [2024]