Engineering Mechanics

Chapter 17 : Linear Motion „„„„„ 367

and total distance travelled by the Train B,

sB= 500 + 20 (T – 50) m ...(ii)
For the train B to overtake the train A, the two distances (i) and (ii) should be equal. Threfore
equating equations (i) and (ii),

390.6 + 12.5 [(T + 60) – 62.5] = 500 + 20 (T – 50)

12.5 T – 31.3 = 109.4 + 20 T – 1000

7.5 T= 1000 – 109.4 – 31.3 = 859.3

∴ 859.3 114.6 s

7.5

T==^ Ans.

EXERCISE 17.1

1. A body starts with a velocity of 3 m/s and moves in a straight line with a constant accelera-
   tion. If its velocity at the end of 5 seconds is 5.5 m/s, find (i) the uniform acceleration, and
   (ii) distance travelled in 10 seconds. [Ans. 0.5 m/s^2 ; 55 m]


2. Two cars start off to a race with velocities (u), and (v), and travel in a straight line with a
   uniform accelerations of (α) and (β). If the race ends in a dead beat, prove that the length
   of the race is:

2

2( – ) ( – )

(– )

uvu vβ α

βα

Hint. Dead beat means that the distance travelled by both the cars is the same.

3. A car starts from rest and accelerates uniformly to a speed of 72 km.p.h. over a distance of
   500 m. Find acceleration of the car and time taken to attain this speed.
   If a further acceleration rises the speed to 90 km.p.h. in 10 seconds, find the new acceleration
   and the further distance moved. [Ans. 0.4 m/s^2 ; 50 sec ; 0.5 m/s^2 ; 225 m ; 62.5 m]


4. A bullet moving at the rate of 300 m/s is fired into a thick target and penetrates up to 500
   mm. If it is fired into a 250 mm thick target, find the velocity of emergence. Take the
   resistance to be uniform in both the cases. [Ans. 212.1 m/s]

17.4. MOTION UNDER FORCE OF GRAVITY

It is a particular case of motion, under a constant acceleration of (g) where its *value is taken
as 9.8 m/s^2. If there is a free fall under gravity, the expressions for velocity and distance travelled
in terms of initial velocity, time and gravity acceleration will be :

1. v = u + gt

2.

(^12)
2
sut gt=+

3. v^2 = u^2 + 2 gs
   But, if the motion takes place against the force of gravity, i.e., the particle is projected up-
   wards, the corresponding equations will be :


4. v = – u + gt

2.

-^12
2

sutgt=+

3. v^2 = – u^2 + 2 gs

* Strictly speaking, the value of g varies from 9.77 m/s^2 to 9.83 m/s^2 over the world. Its value, until and

unless mentioned, is taken as 9.8 m/s^2.