Question

A train 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.

Answer 1

\[60 \cos (15 {}^{\circ})=15 \sqrt{2} \left(1+\sqrt{3}\right)=\frac{57.9555 m}{\sec } \]

Answer 2

The "train is moving at a constant speed", therefore, the speed is constant every where.