Acar is traveling at 120 km/h (75 mph). when applied the braking system can stop the car with a deceleration rate of 9.0 m/s2. the typical reaction time for an alert driver is 0.5 s versus 2 s for a sleepy driver. assuming a typical car length of 5 m, calculate the number of additional car lengths it takes the sleepy driver to stop compared to the alert driver.

a. 13
b. 3.0
c. 10
d. 16
e. 26

c. 10

Explanation:

Both drivers, once applied the brakes, will come to a stop after driving the same distance, which we will call braking distance.

We can get this value (assuming a constant deceleration) applying one of the kinematic equations, as follows:

Vf² -V₀² = 2*a*xf (1)

If the car is stopped completely, then vf=0.

As we know that the deceleration rate is 9.0 m/s², this means it must have a negative sign in the equation above, as it opposes to the movement.

As we have the value of v₀, in km/h, in order to be consistent, it is convenient to convert this units to m/s, as follows:

v₀ = 120 km/h* (1h/3600 sec)*(1000m / 1 km) = 33.3 m/s

Replacing in (1) and solving for xf, we get:

xf = v₀² / 2*a = (33.3 m/s)² / 2* 9.0 m/s² = 61.6 m

Now, in order to get the total distance travelled by both drivers, we need to add the distance that they drove at constant speed, before apply the brakes, which is different for the sleepy driver and  the alert driver.

For the sleepy driver:

xr = 33.3 m/s * 2 s = 66.6 m

⇒ xt = xf + xr = 61.6 m + 66.6 m = 128.2 m

For the alert driver:

xr = 33.3 m/s * 0.5 s = 16.7 m

⇒ xt = xf + xr = 61.6 m + 16.7 m = 78.2 m

The difference between both distances is 50 m, which is equal to 10 additional car lengths for the sleepy driver, compared with the alert driver.

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