Question

For a given particle, a=x^3+2x, then find
distance (x) in terms of time (t), when value of t = 0 and x=0 ?​

Answer 1

answer : -2av³

explanation : The relation between time t and distance x is given by, t = ax² + bx, where a and b are constants.

let's differentiate t with respect to x,

i.e., dt/dx = d(ax² + 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/{dx/dt} = 1/v = 2ax + b

or, v = 1/(2ax + b) ......(2)

now differentiating v with 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 × 2a v

from equation (2),

dv/dt = -v² × 2av = -2av³

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

i.e., A = dv/dt

so, dv/dt = A = -2av³

[here negative sign shows retardation.]

hence, answer is -2av³