Question

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 ?

plzz solve this correctly
wrong answer will be reported
​

Answer 1

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 2

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.