Question

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?

Answer 1

`bb( a) )` 30

`bb( b) )` 40

No units quoted or stated

Explanation:

a)

With respect to time , with `v(t) = 6t`:

`v_("ave") =( int_(bb(Delta t)) \ bb(dt) qquad v(t))/(bb(Delta t))`

`:. v_("ave") =( int_0^10 \ dt qquad 6t)/10`

`= 1/10 [3t^2]_0^10 bb(= 30)`

b)

With respect to distance :

`v_("ave") =( int_(bb(Delta x)) \ bb(dx) qquad v(x))/(bb(Delta x))`

Now:

• `(dv)/(dx) = (dv)/(dt) (dt)/(dx) = 6/v`, variable `t` has been eliminated

Separating out:

`v \ dv = 6 \ dx`

And integrating:

`v^2/2 = 6x + C`

The IV:

• `v(t = 0) = 0, x(t = 0) = 0`

• `implies v(x = 0) = 0 implies C = 0`

So: `bb( v = sqrt(12x) )`

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

• `x(t = 10) = 300`

`:. v_("ave") =( int_0^300 \ dx qquad sqrt(12x))/(300) qquad triangle`

ASIDE :

• `int \ dx \ sqrt(12x) = int \ d(u/12) \ sqrt(u)`

`= 1/12 (u)^(3/2)/(3/2) + C= 1/18 (u)^(3/2) + C`

Evaluating:

`implies triangle = 1/300 ( (12x)^(3/2)/18 )_0^300`

`bb( = 40)`