select the correct answer.\na train is moving at a constant speed on a surface inclined upward at 15.0° with…

Explanation:

Step1: Calculate the speed on the inclined - plane

The speed on the inclined - plane $v=\frac{d}{t}$, where $d = 300$ m and $t = 5$ s. So $v=\frac{300}{5}=60$ m/s.

Step2: Find the horizontal component of the velocity

The horizontal component of the velocity $v_x=v\cos\theta$, where $\theta = 15.0^{\circ}$ and $v = 60$ m/s. Using the formula $v_x=60\times\cos(15^{\circ})$. Since $\cos(15^{\circ})=\cos(45^{\circ}- 30^{\circ})=\cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659$. Then $v_x = 60\times0.9659 = 57.95$ m/s. The speed is constant, so the horizontal velocity at the end of 3 seconds is the same as the horizontal velocity at any time during the motion.

Answer:

A. 57.95 meters/second