Question

the displacement of a body is given to be proportional to the cube of time elapsed what us the nature of the acceleration of the body. justify your answer

Answer 1

let displacement be s
s is prop to t³
s=kt³
on differentiation
ds/dt=3kt²⇒v =3kt²
on again differentiation
d²s/dt=6kt⇒a =6kt
a is prop to t
⇒ acceleration is directly prop to time,it varies linearly with time and increases with resp to time
i hope it help u

Answer 2

Answer:

The acceleration is related to time, fluctuates linearly with time, and increases as time passes.

Explanation:

As per the question,

$x\propto t^3$

$x=kt^3$     (1)

Where,

x=displacement of a body

k=proportionality constant

t=time elapsed

On differentiating equation (1),

$\frac{dx}{dt} =\frac{d(kt^3)}{dt}$

$\frac{dx}{dt} =3kt^2$     (2)

Now on differentiating equation (2) we get;

$\frac{d^2x}{dt^2} =\frac{d(3kt^2)}{dt}$

$\frac{d^2x}{dt^2} =6kt$      (3)

Also,

$a=\frac{d^2x}{dt^2}$        (4)

a=acceleration of the body

On equating equations (3) and (4) we get;

$a=6kt$

$a\propto t$     (5)   (we are considering "6k" as constant)

Hence, acceleration is directly proportional to time, it varies linearly with time and increases with respect to time.  

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