Engineering Mechanics

Chapter 17 : Linear Motion „„„„„ 381

If the train had moved uniformly with a velocity of 30 km/hr, then the time required to cover
1.5 km
60 3
3min
30 2

=×= ...(iii)

∴ Time lost = 4 – 3 = 1 min Ans.
Example 17.33. A cage goes down a main shaft 750 m deep, in 45 s. For the first quarter
of the distance only, the speed is being uniformly accelerated and during the last quarter uniformly
retarded, the acceleration and retardation being equal. Find the uniform speed of the cage, while
traversing the central portion of the shaft.

Solution. Let OABC represent the velocity-time
graph in which OA represents the period of acceleration, AB
the period of uniform velocity and BC the period of

retardation as shown in Fig. 17.8.

First of all consider motion of cage from O to A. In

this case, initial velocity (u 1 ) = 0 (because it goes down

from rest) and distance travelled (s 1 )
750
187.5 m
4

==

Let a 1 = Constant acceleration of the cage, and

v 1 = Uniform velocity of the cage (AD or BE).

We know that area of triangle OAD (s 1 ),

11

1

187.5

2

=××tv

∴ t 1 × v 1 = 187.5 × 2 = 375 ...(i)
Now consider the motion of the cage from A to B. In this case, distance travelled (s 2 )
= 750 – (2 × 187.5) = 375 m

We also know that the area of the rectangle ABED (s 2 ),
375 = t 2 × v 1 ...(ii)
From equation (i) and (ii), we find that
t 1 = t 2
Similarly t 2 = t 3
∴ t 1 = t 2 = t 3
Since the total time taken (t 1 + t 2 + t 3 ) is 45 seconds, therefore
t 1 = t 2 = t 3 = 15 s
Again consider the motion of the cage from O to A. In this case, Initial velocity (u 1 ) = 0 ;
distance travelled (s 1 ) = 187.5 m and time OABC(t 1 ) = 15 s
We know that the distance (s 1 ),
22
11 11 1 1

11

187.5 0 (15) 112.5

22

=+ut at=+a =a

∴

2

1

187.5

1.67 m/s

112.5

a ==

and speed of the cage while traversing the central portion of the shaft

v 1 = u + a 1 t 1 = 0 + 1.67 × 15 = 25 m/s Ans.

Fig. 17.8.