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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.

J. Doe
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  • Isn't it obvious? The second equation doesn't have acceleration so you're not including that. –  Jun 05 '16 at 22:18
  • Yeah, as @ZacharySelk says, you're simply ignoring the acceleration all together in Formula #2. Also, the acceleration should be negative, since it decreases the positive velocity – Bobson Dugnutt Jun 05 '16 at 22:19
  • @Lovsovs is it wrong to use one of these formulas for this problem? There has to be a mistake because I get two different answers when calculating Δx. – J. Doe Jun 05 '16 at 22:23
  • @J.Doe The second formula only works when velocity is constant (which it is not in this case). For this problem, you could multiply the average velocity by time if you wanted. – Hrhm Jun 05 '16 at 22:55

1 Answers1

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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.

Merkh
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