# 1012 Train at constant deceleration | Rectilinear Translation

**Problem 1012**

A train moving with constant acceleration travels 24 ft (7.32 m) during the 10^{th} sec of its motion and 18 ft (5.49 m) during the 12^{th} sec of its motion. Find its initial velocity and its constant acceleration.

**Solution in English Units**

$24 = v_o + 10a$ → equation (1)

$18 = v_o + 12a$ → equation (2)

Equation (1) minus equation (2)

$6 = -2a$

$a = -3 \, \text{ ft/sec}^2$ *answer*

From equation (1)

$24 = v_o + 10(-3)$

$v_o = 54 \, \text{ ft/sec}$ *answer*

**Solution in SI Units**

$7.32 = v_o + 10a$ → equation (1)

$5.49 = v_o + 12a$ → equation (2)

Equation (1) minus equation (2)

$1.83 = -2a$

$a = -0.915 \, \text{ m/sec}^2$ *answer*

From equation (1)

$7.32 = v_o + 10(-0.915)$

$v_o = 16.47 \, \text{ m/sec}$ *answer*