Question

Q. If a(acceleration) = x², then find v(velocity) when x is 2m. Given at x = 0 and body is at rest.

Answer 1

Acceleration and displacement are related as a = x²
we know, acceleration , a = $\bf{v\frac{dv}{dx}}$
so, x² = $\bf{v\frac{dv}{dx}}$
⇒x²dx = vdv
⇒∫x²dx = ∫vdv

at x = 0 , body is in rest ∴ initial velocity , v₀ = 0
Let velocity at x = 2 is v
Now, $\bf{\int\limits^{2}_0{x^2}\,dx=\int\limits^v_0{v}\,dv}$
⇒ [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

Answer 2

Explanation:

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