Case I (max speed = 60km/hr)
a) Acceleration to reach the max speed:
u = 0m/s; v = 60km/hr = 16.67m/s; t = ?;
s = 1km = 1000m; a = ?
v2 = u2 + 2as
16.62 = 02 + 2 (a) (1000)
Therefore,
a = 0.13m/s
v = u + at
16.6 = 0 + 0.13 (t)
Therefore,
t = 127.6 sec
b) Declaration at the end of the station:
u = 60km/hr = 16.67m/s; v = 0m/s; t = ?;
s = 0.5km = 500m; a = ?
v2 = u2 + 2as
02 = 16.672 + 2 (a) (500)
Therefore,
a = - 0.27m/s
v = u + at
0 = 16.67 + (-0.27) (t)
Therefore,
t = 60.18 sec
c) Total time taken
Let the distance between the villages be x meters
Therefore, total time taken
T1 = 127.6 + 60.18 + (x - 1000 ? 500) / 16.67
T1 = 187.78 + (x - 1000 ? 500) /16.67
Case 2 (max speed = 20km/hr)
Follow the same procedure,
a) Acceleration to reach the max speed:
b) Declaration at the end of the station:
c) Total time taken
Then when you get T2
Just subtract T1 from T2, to get the final time elapsed.
Now I don?t know how you got (2 min 40 sec) without (x). Some information is also lacking in Case 2. (ie. the distance covered to reach 20km/hr and declarate)
