The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body, is

Question

The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body, is (A) constant but not zero (B) increasing with time (C) zero (D) decreasing with time

Tardigrade Admin · Accepted Answer

Given that displacement of the body, y propto t3 or y = kt3 ∴ v = (dy/dt ) = 3kt2 or a = (dv/dt) = 6 kt propto t ∴ The magnitude of the acceleration of the bodyincreases with time.

Q. The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body, is

constant but not zero

increasing with time

zero

decreasing with time

Given that displacement of the body,
$y \propto t^{3}$
or $y = k t^{3}$
$∴ v = \frac{d y}{d t} = 3 k t^{2}$
or $a = \frac{d v}{d t} = 6 k t \propto$ t
$∴$ The magnitude of the acceleration of the bodyincreases with time.

Q. The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body, is

constant but not zero

increasing with time

zero

decreasing with time

Given that displacement of the body, y∝t3 ory=kt3 ∴v=dtdy​=3kt2 or a=dtdv​=6kt∝ t ∴ The magnitude of the acceleration of the bodyincreases with time.

Given that displacement of the body,
$y \propto t^{3}$
or $y = k t^{3}$
$∴ v = \frac{d y}{d t} = 3 k t^{2}$
or $a = \frac{d v}{d t} = 6 k t \propto$ t
$∴$ The magnitude of the acceleration of the bodyincreases with time.