Question

The relation between time t and distance x is t = αx² +βx
where α and β are constants. The retardation is
(a) 2αv³ (b) 2βv³ (c) 2αβv³ (d) 2β²v³

Answer 1

Answer:

-2av³

The relation between time t and distance x is given by, t = αx² + βx, where a and b are constants.

let's differentiate t with respect to x,

i.e., dt/dx = d(αx² + βx)/dx

or, dt/dx = 2αx + β .....(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/{dx/dt} = 1/v = 2αx + β

or, v = 1/(2αx + β) ......(2)

now differentiating v with respect to time, t

dv/dt = d{1/(2αx + b)}/dt

= -1/(2αx + β)² × d(2αx + β)/dt

= -1/(2αx + β)² × [2α × dx/dt ]

= -1/(2αx + β)² × 2αv

from equation (2),

dv/dt = -v² × 2αv = -2αv³

we know, acceleration/retardation is the rate of change of velocity with respect to time.

i.e., A = dv/dt

so, dv/dt = A = -2αv³

[here negative sign shows retardation.]

Hence, the answer is -2αv³