Question

The relation between time(t) and distance(x) is t=Ax^2 + Bx, A and B are constants. Prove that its retardation is 2Av^3.

Answer 1

t=Ax^2+Bx
differentiating it with respect to x , we get
1=(2Ax+B) dx/dt
1(/2Ax+B)=dx/dt =v -------(1)
again differentiating it w.r.t x , we gat
-2A/(2A+B)^2=d2x/dt2
from (1)
-2A/v^2=d2x/dt2
hence retardation is 2A/v2

Answer 2

Answer:

Explanation:

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