Question

Suppose a car is traveling at 18.6 m/s, and the driver sees a traffic light turn red. After 0.500 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 4.00 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:

s = 52.545 m

Explanation:

First, we calculate the distance covered during the 0.5 s when the driver notices the light and applies the brake.

$s_{1} = vt$

where,

s₁ = distance covered between noticing light and applying brake = ?

v = speed = 18.6 m/s

t = time = 0.5 s

Therefore,

$s_{1} = (18.6\ m/s)(0.5\ s)\\s_{1} = 9.3\ m$

Now, we calculate the distance for the car to stop after the application of brakes. For that we use 3rd equation of motion:

$2as_{2} = V_{f}^{2} - V_{i}^{2}$

where,

s₂ = distance covered after applying brake = ?

a = deceleration = - 4 m/s²

Vf = final speed = 0 m/s

Vi = initial speed = 18.6 m/s

Therefore,

$2(- 4\ m/s^{2})s_{2} = (0\ m/s)^{2} - (18.6\ m/s)^{2}\\\\s_{2} = \frac{(18.6\ m/s)^{2})}{8\ m/s^{2}}\\\\s_{2} = 43.245\ m$

So the total distance covered by the car before stopping is:

$s = s_{1} + s_{2}\\s = 9.3\ m + 43.245\ m$

s = 52.545 m