Question

1. The displacement of a body is given to be proportional to the
cube of time elapsed. The magnitude of acceleration of the body is
(a) increasing with time
(c) constant but not zero
(b) decreasing with time
(d) zero​

Answer 1

Explanation:

b] decreasing with the time

Answer 2

Answer:

• The magnitude of acceleration of the body will be Increasing with Time.

Given:

1. The displacement of a body is given to be proportional to the  cube of time elapsed ( S ∝ t³ )

Explanation:

$\rule{300}{1.5}$

From The Question We Know,

$\large\bigstar \; {\boxed{\tt S \propto t^3}}$

Now,

$\longmapsto{\large \tt S = Kt^3}$

$\bold{Here}\begin{cases}\text{S Denotes Displacement} \\ \text{t Denotes Time taken} \\ \text{K Denotes Constant}\end{cases}$

$\longmapsto{\large \tt S = Kt^3}$

Differentiating the Equation to get Velocity.

$\longmapsto \large\boxed{\tt v = \dfrac{ds}{dt}}$

$\longmapsto \large{\tt v = \dfrac{d(K \; t^3)}{dt}}$

As K is a constant and cannot be differentiated

$\longmapsto \large{\tt v = K \; \dfrac{d(t^3)}{dt}}$

$\longmapsto \large{\tt v = K \times 3 \; t^2}$

$\longmapsto \large{\underline{\boxed{\tt v = 3 \; Kt^2 \; m/s}}}$

$\rule{300}{1.5}$

$\longmapsto \large{\tt v = 3 \; Kt^2}$

Again, Differentiating the Equation to get Acceleration.

$\longmapsto \large\boxed{\tt a = \dfrac{dv}{dt}}$

$\longmapsto \large{\tt a = \dfrac{d(3 \; K \; t^2)}{dt}}$

As K & 3 is a constant and cannot be differentiated,

$\longmapsto \large{\tt a = 3 \; K \; \dfrac{d(t^2)}{dt}}$

$\longmapsto \large{\tt a = K \times 3 \times 2 t}$

$\longmapsto \large{\underline{\boxed{\tt a = 6 \; Kt \; m/s^2}}}$

From this we can Conclude that,

$\longmapsto \large{\underline{\boxed{\red{\tt a \propto t}}}}$

∵ Acceleration is Directly Proportional to Time Elapsed.

So, the Graph will be a Straight Line.

∴ The magnitude of acceleration of the body increasing with time.

$\rule{300}{1.5}$