(^380) „„„„„ A Textbook of Engineering Mechanics
Fig. 17.7.
Now consider motion of the lift from A to B. We know that area of the triangle ALB (s 2 )
2
1
300
2
=××tv ...(ii)
Dividing equation (i) by (ii),
1
12
2
900
or 3
300
t
tt
t

But t 1 + t 2 = 30 s
∴ t 1 = 22.5 s and t 2 = 7.5 s
Now substituting the value of t 1 in equation (i),
1
900 22.5 11.25
2
=× ×=vv
∴
900
80 m/s
11.25
v== Ans.
(ii) Acceleration of the lift
From geometry of the figure, we find that acceleration, of the lift,
2
1
80
tan 3.55 m/s
22.5
AL
a
OL
=α= = = Ans.
(iii) Retardation of the lift
We also know that retardation of the lift,
a 2 = 3a 1 = 3 × 3.55 = 10.65 m/s^2 Ans.
Example 17.32. A train moving with a velocity of 30 km.p.h. has to slow down to 15 km.p.h.
due to repairs along the road. If the distance covered during retardation be one kilometer and that
covered during acceleration be half a kilometer, find the time lost in the journey.
Solution. Let OABCD be the velocity-time graph, in which AB represents the period of
retardation and BC period of acceleration as shown in Fig. 17.7.
First of all, consider motion of the train from A to B. In this case, distance travelled (s 1 ) = 1 km;
initial velocity (u 1 ) = 30 km. p.h. and final velocity (v 1 ) = 15 km.p.h.
Let t 1 = Time taken by the train to move from A to B.
We know that the area of the trapezium OABE (s 1 )
11
30 15
122.5
2
tt

• =×=
  ∴
  1
  1
  hr 2.67 min
  22.5
  t ==
  ...(i)
  Now consider motion of the train from B to C.
  In this case, distance travelled (s 2 ) = 0.5 km ; Initial velocity
  (u 2 ) = 15 km.p.h. and final velocity (v 2 ) = 30 km.p.h.
  Let t 2 = Time taken by the train to
  move from B to C.
  We also know that the area of trapezium BCDE (s 2 ),
  2
  11530
  22.5
  22
  tt


• 
  

  =×=
  or t 2 =
  1
  45
  hr = 1.33 min ...(ii)
  ∴ Total time, t = t 1 + t 2 = 2.67 + 1.33 = 4 min