Question

Using the distance -time table you have made, compute the average speed of the train between every two consecutive stations.

Answer 1

Answer:

a) 28.7 m/s

b) 25.04 m/s

c) 21.19 m/s

d) 23.37 m/s

e) 25.71 m/s

f) 10.98 m/s

Explanation:

a) S = 465 km = 465000 m

T = 4 hour 30 min = 4×60×60 + 30×60 = 16200 sec

V = S / T

V = 465000 / 16200

V = 28.7 m / sec

b) S = 266000 m

T = 2×60×60 + 57×60 = 10620 sec

V = S / T

V = 266000 / 10620

V = 25.04 m / sec

c) S = 262000 m

T = 3×60×60 + 26×60 = 12360 sec

V = S / T

V = 262000 / 12360

V = 21.19 m/ sec

d) S = 129000 m

T = 1×60×60 + 32×60 = 5520 sec

V = S / T

V = 129000 / 5520

V = 23.37 m / sec

e) S = 233000 m

T = 2×60×60 + 31×60 = 9060 sec

V = S / T

V = 233000 / 9060

V = 25.71 m / sec

f) S = 29000 m

T = 44×60 = 2640 sec

V = S / T

V = 29000 / 2640

V = 10.98 m / sec