Question

If a(accerlaton)=x^2 find vellocity when x is 2m ,Given at x=0,body is at rest

Answer 1

Answer:

Acceleration and displacement are related as a = x²

we know, acceleration , a = \bf{v\frac{dv}{dx}}v

dx

dv

so, x² = \bf{v\frac{dv}{dx}}v

dx

dv

⇒x²dx = vdv

⇒∫x²dx = ∫vdv

at x = 0 , body is in rest ∴ initial velocity , v₀ = 0

Let velocity at x = 2 is v

Now, \bf{\int\limits^{2}_0{x^2}\,dx=\int\limits^v_0{v}\,dv}

0

∫

2

x

2

dx=

0

∫

v

vdv

⇒ [x³/3]²₀ = \bf{[\frac{v^2}{2}]^v_0

⇒8/3 = v²/2

⇒16/3 = v²

So, v = ±4/√3 m/s

Hence, magnitude of velocity = 4/√3 m/s