Question

5. A particle starts moving along x-axis from origin
with initial velocity of 3 m/s. The acceleration a of
the particle varies with position x as a = Ky, where
K is a constant. At x = 3 m, the tangent to the
graph makes an angle of 60° with x-axis. Then at
x = 3 m (v represents velocity of particle at
position x)
1a (m/s)
60°:
x (m)
01
(1) v = 3/2 m/s
(2) v = (9+ (3) m/s
(3) a = 3 m/s2
(4) a = 1.5 m/s2​

Answer 1

Given:

A particle starts moving along x-axis from origin with initial velocity of 3 m/s. The acceleration a of the particle varies with position x as a = kx, where k is a constant. At x = 3 m, the tangent to the graph makes an angle of 60° with x-axis.

To find:

Value of velocity at x = 3 m

Calculation:

The acceleration - displacement function is :

$a = kx$

$= > \dfrac{da}{dx} = k \times \dfrac{dx}{dx} = k$

At x = 3 , the tangent is at 60° ;

$= > \dfrac{da}{dx} = k = \tan(60 \degree)$

$= > k = \sqrt{3}$

Now , the original function is :

$a = kx$

$= > v \dfrac{dv}{dx} = kx$

$= > v \: dv = \sqrt{3} \times x \: dx$

Integrating and putting the limits:

$\displaystyle = > \int_{3}^{v} v \: dv = \sqrt{3} \times \int_{0}^{3} x \: dx$

$= > \dfrac{1}{2} \{ {v}^{2} - {3}^{2} \} = \dfrac{ \sqrt{3} }{2} \{3 - 0 \}$

$= > {v}^{2} - 9 = 3 \sqrt{3}$

$= > {v}^{2} = 3 \sqrt{3} + 9$

$= > v = \sqrt{3 \sqrt{3} + 9}$

So final answer is :

$\boxed{ \sf{ v = \sqrt{3 \sqrt{3} + 9} \: m {s}^{ - 1} }}$