# A particle moves with velocity v=6t from time t=0 to t=10. How do you find the average velocity with respect to a) time and b) distance?

Jun 14, 2018

bb( a) ) 30

bb( b) ) 40

No units quoted or stated

#### Explanation:

a)

With respect to time , with $v \left(t\right) = 6 t$:

${v}_{\text{ave}} = \frac{{\int}_{\boldsymbol{\Delta t}} \setminus \boldsymbol{\mathrm{dt}} q \quad v \left(t\right)}{\boldsymbol{\Delta t}}$

$\therefore {v}_{\text{ave}} = \frac{{\int}_{0}^{10} \setminus \mathrm{dt} q \quad 6 t}{10}$

$= \frac{1}{10} {\left[3 {t}^{2}\right]}_{0}^{10} \boldsymbol{= 30}$

b)

With respect to distance :

${v}_{\text{ave}} = \frac{{\int}_{\boldsymbol{\Delta x}} \setminus \boldsymbol{\mathrm{dx}} q \quad v \left(x\right)}{\boldsymbol{\Delta x}}$

Now:

• $\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dv}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{6}{v}$, variable $t$ has been eliminated

Separating out:

$v \setminus \mathrm{dv} = 6 \setminus \mathrm{dx}$

And integrating:

${v}^{2} / 2 = 6 x + C$

The IV:

• $v \left(t = 0\right) = 0 , x \left(t = 0\right) = 0$

• $\implies v \left(x = 0\right) = 0 \implies C = 0$

So: $\boldsymbol{v = \sqrt{12 x}}$

And, eg, from the $v - t$ graph:

• $x \left(t = 10\right) = 300$

$\therefore {v}_{\text{ave}} = \frac{{\int}_{0}^{300} \setminus \mathrm{dx} q \quad \sqrt{12 x}}{300} q \quad \triangle$

ASIDE :

• $\int \setminus \mathrm{dx} \setminus \sqrt{12 x} = \int \setminus d \left(\frac{u}{12}\right) \setminus \sqrt{u}$

$= \frac{1}{12} {\left(u\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C = \frac{1}{18} {\left(u\right)}^{\frac{3}{2}} + C$

Evaluating:

$\implies \triangle = \frac{1}{300} {\left({\left(12 x\right)}^{\frac{3}{2}} / 18\right)}_{0}^{300}$

$\boldsymbol{= 40}$