Question

12. The nature of graph drawn between displacement in tth
second and time t of a uniformly accelerated motion is
(a) a straight line passing through origin
(b) a straight line with negative y-axis intercept
(c) a parabola
(d) an ellipse​

Answer 1

Answer:

(a)

Explanation:

if we take the distance units in tth

then acceleration units also change to tth/s square,so Nature of the graph dose not change

Answer 2

Given that,

The nature of graph drawn between displacement and time of a uniformly accelerated motion .

Let at initially,

t = 0. x = 0 then, v = 0

Acceleration :

The rate of change of velocity of the object.

$\dfrac{dv}{dt}=a$...(I)

Velocity :

The rate of change of potion of the object.

$v =\dfrac{dx}{dt}$...(II)

Uniform acceleration :

Uniform acceleration means acceleration is constant of the object.

$a = k$

Here, k = constant

We need to calculate the nature of graph between displacement and time.

Using equation (I)

$\dfrac{dv}{dt}=a$

Put the value of acceleration

$\dfrac{dv}{dt}=k$

On integration

$\int_{0}^{v}{dv}=\int_{0}^{t}{dt}$

$v=kt$

Put the value of v from equation (II)

$\dfrac{dx}{dt}=kt$

$dx=kt dt$

On integrating

$\int_{0}^{x}{dx}=k\int_{0}^{t}{t}$

$x=k\dfrac{t^2}{2}$

This equation show the parabolic graph.

Hence, The nature of graph drawn between displacement and time of a uniformly accelerated motion is a parabola.

(c) is correct option