To determine the acceleration of the particle given the relation between time (t) and position (x) as t = αx2 + βx, we need to perform the following steps:
Differentiate with respect to x:
The given equation is t = αx2 + βx.
Differentiate both sides with respect to x to find the expression for dt/dx:
dt/dx = 2αx + β
Calculate velocity:
Velocity (v) is defined as dx/dt. Using the result from step 1, we can write:
v = 1/(dt/dx) = 1/(2αx + β)
Differentiate velocity to find acceleration:
Acceleration (a) is the derivative of velocity with respect to time, a = dv/dt. Using the chain rule:
dv/dt = (dv/dx) * (dx/dt)
Now, first calculate dv/dx:
v = (2αx + β)-1
Differentiate v with respect to x:
dv/dx = -1 * (2αx + β)-2 * 2α
Plug this back into the chain rule expression:
dv/dt = (-2α/(2αx + β)2) * (dx/dt)
Simplify using v:
Recall that v = 1/(2αx + β), so dx/dt = v.
Substitute dx/dt in the expression:
dv/dt = -2αv/(2αx + β)2
Since v = 1/(2αx + β),
(2αx + β) = 1/v
Thus, (2αx + β)2 = 1/v2
Now,
dv/dt = -2αv * v2 = -2αv3
Therefore, the acceleration of the particle is -2αv3.