Question

Review Interactive Solution 2.33 in preparation for this problem. Suppose a car is traveling at +23.8 m/s, and the driver sees a traffic light turn red. After 0.536 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 6.00 m/s2. What is the stopping distance of the car, as measured from the point where the driver first notices the red light?

Answer 1

Answer:

60.0 m

Explanation:

First of all, we need to calculate the distance travelled by the car during the reaction time of the driver. During this time, the car is travelling at constant velocity of

$u=+23.8 m/s$

And the time interval is

$t_1 = 0.536 s$

So, the distance covered is

$d_1 = ut_1 = (23.8)(0.536)=12.8 m$

In the second part, the car decelerates from

u = +23.8 m/s

to

v = 0

Since it comes to a stop. The deceleration is

$a=-6.00 m/s^2$

So we can use the following SUVAT equation to find the distance travelled in this second part:

$v^2-u^2=2ad_2\\d_2 = \frac{v^2-u^2}{2a}=\frac{0-(23.8)^2}{2(-6.00)}=47.2 m$

So, the total distance travelled by the car is

$d=d_1+d_2=12.8+47.2=60.0 m$