E-StoreIrodov is not difficult 2
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.
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