Asked by Rayboy dml
The speed of a goods truck which has been shunted on to a level siding falls from 10km/h to 5km/h in moving a distance of 30m. If the retardation is constant, how much further will the truck travel before coming to to rest?
Answers
Answered by
Anonymous
10 km/h * 1000/3600 = 2.78 m/s
5 km/h = 1.39 m/s
v = V i + a t
1.39 = 2.78 + a t
so
a = -1.39/t
x = Xi + Vi t + (1/2) a t^2
30 = 0 + 2.78 t - 0.694 t = 2.09 t
t = 14.4 seconds
a = -1.39/14.4 = - 0.0965 m/s^2
x = average speed during stop * time to stop
average speed = 1.39/2= 0.695 m/s
0 = 1.39 - 0.0965t
t = 14.4 s
x = 0.695 * 14.4
= 10 meters
5 km/h = 1.39 m/s
v = V i + a t
1.39 = 2.78 + a t
so
a = -1.39/t
x = Xi + Vi t + (1/2) a t^2
30 = 0 + 2.78 t - 0.694 t = 2.09 t
t = 14.4 seconds
a = -1.39/14.4 = - 0.0965 m/s^2
x = average speed during stop * time to stop
average speed = 1.39/2= 0.695 m/s
0 = 1.39 - 0.0965t
t = 14.4 s
x = 0.695 * 14.4
= 10 meters
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