# The reaction time for an automobile driver is 0.7 sec.If the automobile can be decelarated at 5 m/s^2 calculate the total distance travelled in coming to stop from an initial velocity of 8.33 m/s after a signal is observed?

Jun 20, 2018

The car will move 13.60 m

#### Explanation:

We have distance $s = v t$
${v}_{1} = {v}_{0} + a t$
Before the driver reacts, the car will have moved
${s}_{1} = 8.33 \cdot 0.7 m = 5.831 m$
The time for the car to stop can be found from
${v}_{1} = {v}_{0} + a t$
$0 = 9.33 - 5 t$ or t=1.866 s
The average speed during this time is ${v}_{a} = \frac{1}{2} \cdot 8.33 \frac{m}{s} = 4.165 \frac{m}{s}$
Therefore, the car moves $4.165 \cdot 1.866$ m =7.772m

Total distance moved before the car stops, therefore, is
(5.831+7.772) m = 13.603 m
Rounded to 2 decimals: 13.60 m