# Question #d6159

##### 1 Answer

There is no way one can find the exact value of velocity at

#### Explanation:

At the best one can find an approximate value of velocity at

Notice that value obtained for

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Alternative method.

We need to find kinematic expression represented by the given graph.

Let it be

#s(t)=s_0+ut+1/2at^2# ......(1)

To ascertain three unknown. We are required to have three equations. Choosing three points where both

- At
#t=0# , (1) becomes

#s(0)=s_0+uxx0+1/2axx0^2#

#=>s(0)=s_0#

From the graph#s(0)=25m# . Hence,

#s_0=25m# .......(2) - At
#t=4# , (1) becomes

#s(4)=25+uxx4+1/2axx4^2#

Again from the given graph

#21=25+uxx4+1/2axx4^2#

#=>4u+8a=-4#

#=>a=-0.5(1+u)# .......(3) -
- At
#t=6# , (1) becomes

#s(6)=25+uxx6+1/2axx6^2#

Again from the given graph

#16=25+6u+18a#

#=>6u+18a=-9#

#=>2u+9a=-3#

Using (3) we get

#2u+6(-0.5(1+u))=-3#

#=>2u-3u=-3+3#

#=>u=0ms^-1# .......(4)

Inserting this value in (3) we get

#a=-0.5ms^-2# .........(5)

- At

Using (2), (4) and (5) equation (1) becomes

Graph can be represented as this equation.

We know that velocity

To find out velocity, either we differentiate (6) with respect to time

Or we may use the following kinematic expression

#v=u+at#

Using (4) and (5), above equation becomes

.-.-.-.-.-.-.-.-.-.-.-

If you like you may verify equation (6) for other points on the graph. For example