Which formula is appropriate to use when calculating the displacement in this case?

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The "reaction time" of the average automobile driver is about 0.7s. If an automobile can slow down with an acceleration of 12.0 ft/s^2, compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 15.0 mph.


First, I converted the 15.0 mph to feet/second and got 22 ft/s.

Formula #1:
Δx = $v_1$Δt + (1/2)a$Δt^2$
$Δx = 22(0.7) + (1/2)(-12)(0.7^2) = 12.46 ft$

Formula #2:
$Δx = v_1 Δt$
$Δx = (22)(0.7) = 15.3 ft$

I am not sure which formula to use for this problem.

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First you calculate how long before the driver reacts. 22ft/s times 0.7s = feet traveled before the driver reacts. Then, the driver reacts. How long does it take for the driver to completely stop (forgetting about jerk and higher derivatives)? Eyeballing it, about 1.8 seconds. This is the $\Delta t$ you use in formula 1, with $v = 22$ initially.