Answer to Question #266017 in Physics for Gayle

Question

Train starts at 37 m/s to 12 m/s after 5 seconds. What is the acceleration of the train at this time? How far from starting point to the point after 5 seconds does the train travel? (Please show the solution)

Answer 1

The acceleration:

$$a=\frac{v_f-v_i}{t}=\frac{12-37}{5}=-5\text{ m/s}^2.$$

The distance:

$$d=\frac{v_f^2-v_i^2}{2a}=\frac{12^2-37^2}{2(-5)}=122.5\text{ m}.$$