Lefty is driving the getaway car after a bank robbery. He's going 50 m/s when he spots a nail strip 150 m in front of him. If Lefty can stop in time, he can avoid the strip and the cops. If not, Lefty's tires are punctured, he is caught, and spends the next 20 years in Joliet. Lefty's reaction time is 0.6 seconds and his car's maximum deceleration is 10 m/s2.
How far from the nail strip does he stop? Enter a negative number if he stops before he reaches the strip, or a positive number if he passes the strip and his tires are punctured.

Question

Lefty is driving the getaway car after a bank robbery. He's going 50 m/s when he spots a nail strip 150 m in front of him. If Lefty can stop in time, he can avoid the strip and the cops. If not, Lefty's tires are punctured, he is caught, and spends the next 20 years in Joliet. Lefty's reaction time is 0.6 seconds and his car's maximum deceleration is 10 m/s2.
How far from the nail strip does he stop? Enter a negative number if he stops before he reaches the strip, or a positive number if he passes the strip and his tires are punctured.

anonymous · Answer

Vi = 50 m/s
Vf = 0 m/s
a = -10 m/s^2
Xf = ? m
Xi = 0 m

Vf^2 = Vi^2 + 2a(Xf - Xi)
0 = 50^2 + 2(-10)(Xf - 0)
-2500 = -20 Xf
Xf = 125 m

This means that he will need 125 m to come to a stop. BUT, we need to factor in that he takes 0.6 s initially to react and press the brakes. This means he covers some distance before he hits the brakes.

X = v*t = (50)(0.6) = 30 m

If we sum the distance he went before braking and the distance he went while braking:
Xtot = 30m + 125m = 155 m

Unfortunately for him, he is going to run over the strip. He went over +5 m.