Atrain is moving at a constant speed on a surface inclined upward at 15.0° with the horizontal and travels 300 meters in 5 seconds. calculate the horizontal velocity of the train at the end of 3 seconds

It would be helpful to draw the system to fully understand the situation. We would see that a right triangle can be made. We can use trigonometric functions so as to relate terms.

velocity, v = 300/5 = 60 m/s
horizontal component of velocity = vcos(theta) = 60cos15 = 57.96 m/s

It will have the same horizontal velocity at the end of 3 seconds which is 57.96 m/s since it is moving at a constant speed. Hope this answers the question. Have a nice day.

Hi, your answer is not correct. this is a cumulative