A particle moving in a straight line traverses a distance x in time t....
Explanation:
Given,
Distance travelled, x = x
Time taken, t = x2/2x
To find the acceleration and retardation of the particle, we need to differentiate the distance equation with respect to time.
Differentiating x with respect to time, we get the velocity equation, v.
Differentiating v with respect to time, we get the acceleration equation, a.
Differentiating a with respect to time, we get the retardation equation, r.
Finding Velocity:
We know that,
v = dx/dt
Differentiating x with respect to time, we get
v = x/t
Substituting the value of t from the given equation, we get
v = 2x/x2
v = 2/x
Finding Acceleration:
We know that,
a = dv/dt
Differentiating v with respect to time, we get
a = -2/x2
Finding Retardation:
We know that,
r = da/dt
Differentiating a with respect to time, we get
r = 4/x3
Comparing the options with the derived equations, we can see that Option A is correct.
Therefore, the retardation of the particle is the cube of its velocity.
A particle moving in a straight line traverses a distance x in time t....
We have, t = x2/2 + x
Therefore, dt/dx = 2x/2 + 1 = x + 1
Thus, if v be the velocity of the particle at time t, then
v = dx/dt = 1/(dt/dx)
= 1/(x + 1) = (x + 1)-1
Thus dv/dt = d((x + 1)-1)/dt
= (-1)(x + 1)-2 d(x + 1)/dt
= -1/(x + 1)2 * dx/dt
As, 1/(x + 1) = dx/dt,
So, -(dx/dt)2(dx/dt)
Or dv/dt = -v2*v [as, dx/dt = v]
= -v3
We know, dv/dt = acceleration of a particle.
As, dv/dt is negative, so there is a retardation of the particle.
Thus, the retardation of the particle = -dv/dt = v3 = cube of the particle.