Question

The acceleration of a particle moving along x- axis is a= -100x+50. It is released from x=2. Here a and x are in SI units. The speed of the particle at origin will be......

Answer 1

Answer:

$a = 50 - 100x$

$= > v \dfrac{dv}{dx} = 50 - 100x$

$= > v \: dv = (50 - 100x)dx$

Integrating on both sides :

$= > \int v \: dv = \int \: (50 - 100x)dx$

Applying given limits ( i.e. at x = 2m , the Velocity was 0 , and at x = 0 m , Velocity be v)

$= > \int_{0}^{v} v \: dv = \int_{2}^{0} \: (50 - 100x)dx$

$= > \bigg \{ \dfrac{ {v}^{2} }{2} \bigg \}_{0}^{v} = \bigg \{50x - 50 {x}^{2} \bigg \}_{2}^{0}$

$= > \dfrac{ {v}^{2} }{2} = \{0 - (50 \times 2) \} - \{0 - (50 \times {2}^{2}) \}$

$= > \dfrac{ {v}^{2} }{2} = 200 - 100$

$= > \dfrac{ {v}^{2} }{2} = 100$

$= > {v}^{2} = 200$

$= > v = \sqrt{200}$

$= > v = 10 \sqrt{2} \: m {s}^{ - 1}$

So final answer :

$\boxed{ \boxed{\sf{ \red{ \large{Velocity_{(at \: x = 0 \: m)} = 10 \sqrt{2} \: m {s}^{ - 1} }}}}}$

Answer 2

Answer:

• Velocity (v) of the Particle is 10√2 m/s.

Given:

1. Acceleration (a) = - 100 x + 50
2. Initial Position x = 2 & Final Position x = 0.

Explanation:

$\rule{300}{1.5}$

From the Acceleration we Know,

$\large \bigstar\;{\boxed{\tt a = v.\dfrac{dv}{dx}}}$

$\bold{Here}\begin{cases}\text{v Denotes Velocity} \\ \text{dv Denotes small velocity} \\ \text{dx Denotes small distance Covered in small time}\end{cases}$

Now,

$\large{\boxed{\tt a = v.\dfrac{dv}{dx}}}$

Simplifying,

$\longmapsto\large{\tt a = v.\dfrac{dv}{dx}}$

$\longmapsto\large{\tt a.dx = v.dv}$

$\longmapsto\large{\tt v.dv = a.dx }$

Integrating,

$\longmapsto\large{\tt \displaystyle\int \tt v.dv = \displaystyle\int \tt a.dx }$

Applying limits,

$\longmapsto\large{\tt \displaystyle\int\limits_{0}^{v} \tt v.dv = \displaystyle\int\limits_{2}^{0} \tt a.dx }$

Here lower limit is x = 2 as its is initial position & Upper limit is x = 0 as it is final position.

Now,

$\longmapsto\large{\tt \displaystyle\int\limits_{0}^{v} \tt v.dv = \displaystyle\int\limits_{2}^{0} \tt a.dx }$

$\longmapsto\large{\tt \displaystyle\int\limits_{0}^{v} \tt v.dv = \displaystyle\int\limits_{2}^{0} \tt (-100x + 50).dx }$

$\longmapsto\large{\tt \displaystyle\int\limits_{0}^{v} \tt v.dv = \displaystyle\int\limits_{2}^{0} \tt -100x.dx+ \displaystyle\int\limits_{2}^{0} \tt 50.dx }$

$\longmapsto \large{\tt \Bigg[\dfrac{v^2}{2}\Bigg]^v_0 = \Bigg[\dfrac{-100x^2}{2}\Bigg]^0_2 + \Bigg[50x\Bigg]^0_2}$

$\longmapsto \large{\tt \Bigg[\dfrac{v^2}{2}\Bigg]^v_0 = \Bigg[\cancel{\dfrac{-100x^2}{2}}\Bigg]^0_2 + \Bigg[50x\Bigg]^0_2}$

$\longmapsto \large{\tt \Bigg[\dfrac{v^2}{2}\Bigg]^v_0 = \Bigg[- 50x^2\Bigg]^0_2 + \Bigg[50x\Bigg]^0_2}$

Substituting,

$\longmapsto \large{\tt \dfrac{v^2}{2} = \Bigg[- 50 \times (0)^2 - (-50 \times (2)^2)\Bigg] + \Bigg[50 \times 0 - 50 \times 2\Bigg]}$

$\longmapsto \large{\tt \dfrac{v^2}{2} = \Bigg[0 + 200 \Bigg] + \Bigg[ 0 -100 \Bigg]}$

$\longmapsto \large{\tt \dfrac{v^2}{2} = \Bigg[200 \Bigg] + \Bigg[-100 \Bigg]}$

$\longmapsto\large{\tt \dfrac{v^2}{2} = 200 - 100}$

$\longmapsto\large{\tt \dfrac{v^2}{2} = 100}$

$\longmapsto\large{\tt v^2 = 2 \times 100}$

$\longmapsto\large{\tt v^2 = 200}$

$\longmapsto\large{\tt v = \sqrt{200}}$

$\longmapsto\large{\underline{\boxed{\red{\tt v = 10 \sqrt{2} \; m/s}}}}$

∴ Velocity (v) of the Particle is 10√2 m/s.

$\rule{300}{1.5}$