Engineering Mechanics

(^390) „„„„„ A Textbook of Engineering Mechanics
where C 1 is the first constant of integration. Substituting the values of t = 1 and v = 6.25 in
equation (ii),
11
1
6.25 – 1 5 4.25
4
=++=+CC
∴ C 1 = 6.25 – 4.25 = 2
Substituting this value of C 1 in equation (ii),
4
–52^3
4
t
vtt=++ ...(iii)
Now for velocity of the particle, substituting the value of t = 2 in the above equation ,
4
(2) – (2) (^3) (5 2) 2 8 m/s
4
v=+×+= Ans.
Displacement at t = 2 seconds
Rewriting equation (iii),
4
–52^3
4
ds t
tt
dt
=++ ...
ds
v
dt
⎛⎞
⎜⎟=
⎝⎠
Q
∴
4
–52^3
4
t
ds t t dt
⎛⎞
=++⎜⎟⎜⎟
⎝⎠
...(iv)
Integrating both sides of equation, (iv)
54 2
2
5
–2
20 4 2
tt t
stC=+++ ...(v)
where C 2 is the second constant of integration. Substituting the values of t = 1 and s = 8.8 in
equation (v),
22
115
8.8 – 2 4.3
20 4 2
=+++=+CC
∴ C 2 = 8.8 – 4.3 = 4.5
Substituting this value of C 2 in equation (v) ,
54 2 5
–24.5
20 4 2
tt t
st=+++
Now for displacement of the particle, substituting the value of t = 2 in the above equation,
32 16 20

• 4 4.5 16.1 m
  20 4 2

s=+++=^ Ans.

Example 18.5. A train, starting from rest, is uniformly accelerated. The acceleration at any

instant is

10

v1+

m/s^2 , where (v) is the velocity of the train in m/s at the instant. Find the distance, in

which the train will attain a velocity of 35 km. p.h.

Solution. Given : Equation of acceleration :

10

1

a

v

=

+

Rewriting the given equation,

10

·

1

dv

v

ds v

=

+

... ·

dv

av

ds

⎛⎞

⎜⎟=

⎝⎠

Q

∴ v (v + 1) dv= 10 ds ...(i)

or (v^2 + v) dv= 10 ds