Question

physics class xi
A particle moves along the parabolic path
$2x - {x}^{2} + 2$
In such a way that x component of velocity vector remains constant(5m/s). Find the magnitude of acceleration of particle

Answer 1

Answer:

Explanation:

x component of velocity = 5m/sec

Let Vx be x component of velocity and

Vy be y component of velocity

Let ax be x component of acceleration and

ay be y component of acceleration

Vx = 5m/sec

ax = 0

Vx=dx/dt  Vy=dy/dt

ax=d^2x/dt^2    ay= d^2y/dt^2

$y=2x-x^2+2\\dy/dt=(2-2x)dx/dt\\d^2y/dt^2=(2-2x)d^2x/dt^2 + (-2)dx/dt\\ay=(2-2x)ax-2dx/dt\\ay=-2*5=-10$

(we use differentiaion formula d(a * b)/dx = a * da/dx + b* db/dx )

(and put ax = 0)

ay = -10

ax= 0

Net a = -10 m/sec^2

Hope it helps :-)