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

Task. A train is moving at a constant speed on a surface inclined upward at $15.0{}^{\circ}$ with the horizontal and travels $d=300$ meters in $t=5$ seconds. Calculate the horizontal velocity of the train at the end of 3 seconds.

Solution. The surface is inclined upward at $15.0{}^{\circ}$ with the horizontal, therefore the horizontal velocity of the train at time $t$ is equal to

$v_{hor}(t)=v(t)\cos 15{}^{\circ}.$

By assumption the speed of the train along surface is constant, and so it is equal to

$v=\frac{d}{t}=\frac{300}{5}=60\ m/s.$

Hence

$v_{hor}=v\cos 15{}^{\circ}=60*0.96593=57.95554\approx 58\ m/s$

and it does not depend on $t$. Therefore at the end of 3 seconds, the horizontal velocity will be equal to $58\ m/s$.

Answer. $58\ m/s$.