1012 Train at constant deceleration | Rectilinear Translation

$24 = v_o + 10a$   →   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

$7.32 = v_o + 10a$   →   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