The next  problem be best tackled with graph. Let us take the Q.

Q.1.3

A car starts rectilinearly, first with acceleration w = 5.0 m / s ² ( the initial velocity is equal to zero ) , then uniformly , and finally decelerating at the same rate w , comes to a stop. The total time of motion  equals T = 25 secs. The average velocity during that time is equal to  v = 72 km / hr. How long does the car moves uniformly ?

soln :

The problem is almost, but not exactly similar to problem solved in http://www.goiit.com/posts/list/mechanics-hope-u-people-know-40142.htm#205081

Let us consider time along X axis and velocity along Y axis.

Let the car accelerates from t = 0  up to time t₁ ( st. line with +ve slope ) , then moves with constant velocity from t= t₁  to t = t₂ ( st line || to time axis ) and finally with deceleration (  st. line with  -ve slope ) from t = t₂  to t = T.

our intention is to calculate t ₂ - t₁ = t ₀ ( say )

Now from the graph, ( which I can't draw and show here -- but just like elessar_iitkgp (1933) sir's method  )


Now total area under V-t graph = total distance traversed.  

   S  = ( 1/2) t₁ V₀ +  (t ₂ - t₁ ) V₀   +( 1/2) ( T- t₁ - t₂ ) V₀

    S  = ( 1/2) V₀ ( T + t ₀ )

Total time = T

Remembering,  acceleration w = V₀ / t₁  [ from fig, invisible ] etc,

we get finally, t₀ = T ^{[ ]}  [ 1 - ( 4v /w ) ]

where v is the average velocity = | displacement | / total time

plugging the values,  t₀ = 15 sec.