#### To determine

**To show:**

(i) The average of the velocities with respect to t is vave=12vT

(ii) The average of the velocities with respect to s is vave=23vT

**Solution:**

In the following we prove that

(i) The average of the velocities with respect to t is vave=12vT

(ii) The average of the velocities with respect to s is vave=23vT

#### Explanation

**1) Concept:**

Use the mean value theorem for integral to show the average of the velocities with respect to time t and with respect to displacement s

**2) Mean Value Theorem for Integrals**

If f is continuous on a,b then there exists a number c in a,b such that

fc=fave=1b-a∫abfx dx,

that is,

∫abfx dx=fcb-a

**3) Given:**

If a freely falling body starts from rest, then its displacement is given by

s=12gt2

**4) Calculation:**

st=12gt2 …(1)

Solve the above equation for t.

2s=gt2

2sg=t2

Taking square root,

t=2sgsince t≥0…2

Here, s is the displacement of a freely falling body with respect to the time (t).

The velocity of the body (v) is the rate of change of the displacement (s) with thetime (t).

v=dsdt=ddt12gt2

Differentiate with respect to t.

=12g2t=gt

Therefore,

v=gt

Substitute the value of t from (2).

v=g2sg=2sg

Squaring on both sides,

v2=2sg

Solve the equation for s.

s=v22g

Therefore, v can be a function of t or a function of s.

v=Ft=gt and v=Gs=2gs

vT=FT=gT

Displacement can be viewed as a function of t.

s=st=12gt2

Also since v=Ft=gt,

st=v22g=Ft22g

When t=0

s0=12gt2=12g0=0

After time t=T

sT=FT22g

2gsT=FT2

2gsT=FT2

2gsT=FT·FT

2gsT=gT·FT

2sT=T·FT

Therefore,

vT=gT=FT=2sTT ..3

The average of the velocity with respect to the time t during the interval 0,T is

vt-ave=Fave=1T-0∫0TFtdt

Recall that Ft=gt=s'(t). So by the fundamental theorem of calculus,

1T-0∫0TFtdt=1T[sT-s0]

Since s0=0

1T-0∫0TFtdt=sTT

By 3,

=12vT

Now the average of the velocities with respect to the displacement s during the corresponding interval s0,sT=0,sT is

vt-ave=Gave=1sT-0∫0sTGs ds

=1sT∫0sT2gs ds

=2gsT∫0sTs12 ds

=2gsT·23s320sT

=23·2gsTsT32

=232gsT

Now recall that v(t)=Gs(t)=2g(t)

So,

232gsT=23vT

**Conclusion:**

(i) The average of the velocities with respect to t is vave=12vT

(ii) The average of the velocities with respect to s is vave=23vT