A torin travels at 60 km/h for 0.52h, 30 km/h for next 0.24 h, and then to 70 km/h for the rest .What is the average speed of the train ?
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Answers
Answer:
orin travels at 60 km/h for 0.52h, 30 km/h for next 0.24 h, and then to 70 km/h for the rest .
Answer:
This is an acceleration, time and speed problem where the acceleration is -0.2 m/s^2 and speed is 72 km/h. The distance traveled by the train after applying the brakes until it stops is to be calculated.
First 72 km/h must be converted to m/s. Converting, 72,000 m / 3600 s gives 20 m/s.
The time is not yet known so to solve for it we use t = change in speed / acceleration.
time = (0 m/s - 20 m/s) / -0.2 m/s^2)
time = -20 m/s / -0.2 m/s^2
time = 100 s
Using the formula d = v * t + 1/2 * a * t^2 where d is the distance traveled, a is the acceleration and t is the elapsed time.
d = 20 m/s * 100 s + 1/2 * -0.2 m/s^2 * 100 s^2
d = 2000 m + -1000 m
d = 1000 m
The distance traveled by the train after its brakes were applied is 1000 m.