Question

Q.The velocity of any part:cle is related with its displacement x =
$\sqrt{v + 1}$

Calculate acceleration at x = 5m.
​

Answer 1

Given :

Relation between velocity of particle with its displacement :

$\sf\:x=\sqrt{v+1}$

To Find :

Accelaration at x = 5 m

Theory :

• Velocity

The rate of change of displacement of a particle with time is called velocity of the particle.

${\purple{\boxed{\large{\bold{Velocity=\frac{Distance}{Time\:interval}}}}}}$

In differential form:

$\sf\:Velocity,V=\dfrac{dx}{dt}$

• Acceleration

It is defined as the rate of change of velocity.

${\purple{\boxed{\large{\bold{Acceleration=\frac{Velocity}{Time\:interval}}}}}}$

In differential form:

$\sf\:Acceleration,a=\dfrac{dv}{dt}$

Solution :

We have ,

$\rm\:x=\sqrt{v+1}$

On squaring both sides

$\sf\implies\:x^2=v+1$

$\sf\implies\:v=x^2-1$

Now , Differentiate with respect to t

$\sf\implies\dfrac{dv}{dt}=\dfrac{d(x^2)}{dx}\times\dfrac{dx}{dt}-0$

$\sf\implies\dfrac{dv}{dt}=2x\times\:v-0$

$\sf\implies\dfrac{dv}{dt}=2xv$

$\sf\implies\:a=2x(x^2-1)$

We have , to find Accelaration at x = 5 m

Thus ,

$\sf\implies\:a=2(5)(5^2-1)$

$\sf\implies\:a=10(24)$

$\sf\implies\:a=240ms^{-2}$

$\rule{200}2$

More information about topic

• Both Velocity and Acceleration are vector quantities.
• The velocity of an object can be positive, zero and negative.
• SI unit of Velocity is m/s
• SI unit of Acceleration is m/s²
• Dimension of Velocity: $\sf\:[M^0LT{}^{-1}]$
• Dimension of Acceleration: $\sf\:[M^0LT{}^{-2}]$

Answer 2

Answer:

We have , to find Accelaration at x = 5 m

Thus ,

\sf\implies\:a=2(5)(5^2-1)⟹a=2(5)(5

2

−1)

\sf\implies\:a=10(24)⟹a=10(24)

\sf\implies\:a=240ms^{-2}⟹a=240ms

−2