If e1 = 13.0 v and e2 = 5.0 v , calculate the current i2 flowing in emf source e2.

The solution would be like this for this specific problem:

KCL at Junction a. 
+ I1 + I2 + I3 = 0 (1) 

KVL
+ 13 V - 0.2 R I1 - 0.025 R I1 - 5 V + 0.02 R I2 = 0 (2) 
8 + 0.02 I2 = 0.225 I1 
I1 = 35.6 + 0.0889 I2 (2A) 

KVL (bottom loop - CCW direction) 
- 0.02 R I2 + 5 V + 0.5 R I3 = 0 (3) 
0.5 I3 = -5 + 0.02 I2 
I3 = -10 + 0.04 I2 (3A) 

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s