Question

Velocity v and position x of a body are related as
V2 = kx, where K is a constant. Velocity of the body
after 1 s is (given that at t = 0, x = 0)
(1) K
(2) K/2
(3) 2K
(4) K
​

Answer 1

Explanation:

v2=kx.....

we know..v2-u2=2as

initial velocity is zero......

so v2=2ax......kx=2ax......x cancel

we get k=2a........a=v/t.....as time is 1 sec so k=2v/t......k=2v.......v=k/2

Answer 2

Concept

For a body moving with constant acceleration, the speed of the body at time t is given by

                                             v = u + at.                        

Given

• The speed of the body is related to the position as v² = kx.        ...(1)

• At t = 0, x = 0.

Find

The speed of the body after 1s.

Solution

Acceleration of the body

If the speed of a body varies directly with the distance covered then, the acceleration of the body is constant as we find below.

     v² = kx                        (Using eq (1))

⇒    v = $\sqrt{kx}$.                           ...(2)

Acceleration of the body,

       $a = \frac{dv}{dt} = \frac{\sqrt{k}}{2\sqrt{x}} \cdot \frac{dx}{dt} = \frac{\sqrt{k}}{2\sqrt{x}} \cdot \sqrt{kx}$               (Using eq (2))

⇒     $a$ = k/2.                                            ...(3)

k is a constant so $a$ is also constant.

Speed of the body as a function of time

Since the body is moving with constant acceleration, the velocity of the body is given by the equation of motion,

          v = u + $a$t.                     ...(4)

At t = 0, x = 0. Using equation (2) we get

x = 0, v = 0.

So, at t = 0, v = 0.

Put it in equation (4).

0 = u + 0  ⇒  u = 0.

The initial speed of the body is zero. So, the speed of the body is given by.

       v = at.                ...(5)

Speed of the body at t = 1s

Put t = 1s in eq (5).

v(1s) = $a$ m/s = k/2 m/s.              (Using eq 3)

Speed of the body after 1s is k/2 m/s. So, option 2 is correct.

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