Question

the relation between time t and distance x is t=ax 2 +bx where a and b are constants. express the instantaneous acceleration and instantaneous velocity

Answer 1

The relation between time t and distance x is given by, t = ax^2 + bx where a and b are constants.

Let's differentiate t with respect to x,

i.e., dt/dx = d(ax^2 + bx) /dx

Or, dt/dx = 2ax + b....(1)

We know, Velocity is the rate of change of displacement with respect to time.

i.e.,v = dt/dx

From equation (1),

dt/dx = 1/V = 2ax + b

Or, V = 1/(2ax +b) c....(2)

Now differentiating V with respect to time t,

dv/dt = d{1/(2ax + b)} /dt

= - 1/(2ax +b) ^2 × d(2ax + b) /dt

= - 1/(2ax +b) ^2 × [2a × dx/dt]

= - 1/(2ax +b) ^2 × 2ab

From equation (2),

dv/dt = - v^2 × 2av = - 2av^3

We know, Accerleration /retardation is the rate of change of Velocity with respect of time.

i.e., A = dv/dt

So, dv/dt = A = - 2av^3

[Here negative sign shows retardation.]

Hence, answer is - 2av^3.

Answer 2

Answer:

The relation between time t and distance x is given by, t = ax^2 + bx where a and b are constants. We know, Velocity is the rate of change of displacement with respect to time. We know, Accerleration /retardation is the rate of change of Velocity with respect of time.