Name: Motion With Variable Acceleration
Uploaded: 2019-07-19
Duration: 10 min 54 s
Description: You may be aware of the SUVAT equation, but do you think this can be applicable for all types of motion? No, what are those motion? Watch this to find out.

Question

A particle is projected at an angle $θ$ with an inclined plane making an angle $β$ with the horizontal as shown in the figure, speed of the particle is $u,$ after time $t$ find:

(a) $x$ component of acceleration?
(b) $y$ component of acceleration?
(c) $x$ component of velocity?
(d) $y$ component of velocity?
(e) $x$ component of displacement?
(f) $y$ component of displacement?
(g) $y$ component of velocity when the particle is at maximum distance from the inclined plane?

Accepted Answer

When a particle is projected in air its acceleration is in vertical downward direction and is equal to $g$ .
According to given situation the axes are along and perpendicular to the inclined plane. Components of acceleration and velocity of particle are shown in figure.

$(a)$ From figure, $x -$ component of acceleration, $a_{x} = - g \sin β . . . (1)$
$(b)$ From figure, $y -$ component of acceleration, $a_{y} = - g \cos β . . . (2)$

$(c)$ From figure, $x -$ component of initial velocity, $u_{x} = u \cos θ . . . (3)$
$\Rightarrow$ Using first equation of motion along $x -$ direction, we can say that $x -$ component of velocity $(v_{x})$ at time $t$ is :
$v_{x} = u_{x} + a_{x} t$
$v_{x} = u \cos θ - g \sin β t . . . (4)$
$(d)$ From figure, $y -$ component of initial velocity, $u_{y} = u \sin θ . . . (5)$
$y$ Using first equation of motion along $y -$ direction, we can say that $y -$ component of velocity $(v_{y})$ at time $t$ is :
$v_{y} = u_{y} + a_{y} . t$
$\Rightarrow v_{y} = u \sin θ - g t \cos β . . . (6)$

$(e)$ Using equation $(1), (3)$ and second equation of motion, the displacement along $x$ -axis is given by,
$x = u_{x} t + \frac{1}{2} a_{x} t^{2}$
$\Rightarrow x = u \cos θ t - \frac{1}{2} g t^{2} \sin β$
$(f)$ Using equations $(2), (5)$ and second equation of motion,
$y = u_{y} t + \frac{1}{2} a_{y} t^{2}$
$y = u \sin θ - \frac{1}{2} g t^{2} \cos β$
$(g)$ Particle will move away from the inclined plane till its component of velocity along $y -$ direction is non-zero. So at maximum distance, the velocity is parallel to inclined plane, i.e., $y -$ component of velocity is zero.

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