Question

1. A particle starts from the origin of coordinates at time
t = 0 sec and moves in the X - Y plane with a constant
acceleration a in the Y --direction. Its equation of motion is
$y = \beta {x}^{2}$
.Its velocity component in the X-direction is -
(a) variable
(b) 2a/B
(c) a/2B
(d) a/2B​

Answer 1

Answer:

$vx=\dfrac{a}{2{\beta}}$

Explanation:

Since acceleration is in y-direction and is constant, let's take $ay = \alpha$

thus $\dfrac {dvy}{dt} = \alpha$

Integrate to get, $vy =\alpha {t}+k$

Since the particle starts moving at t=0, thus k=0.

Thus,

$vy =\alpha {t}$

Also, integrate once again to get $y=\dfrac {a{t}^{2}}{2}$ (y=0 at t=0)

now,

$y= \beta {x}^{2}$

hence $x =\sqrt {{{\dfrac {y}{\beta}}}}$

$= \sqrt {\dfrac{{a{t}^{2}}}{2\beta}}$

Differentiate with time to get, $vy= \beta(2x)(vx)$

Solve to get

$vx= \dfrac{a}{2{\beta}}$