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.
[tex]s_{1} = vt\\[/tex]
where,
s₁ = distance covered between noticing light and applying brake = ?
v = speed = 18.6 m/s
t = time = 0.5 s
Therefore,
[tex]s_{1} = (18.6\ m/s)(0.5\ s)\\s_{1} = 9.3\ m\\[/tex]
Now, we calculate the distance for the car to stop after the application of brakes. For that we use 3rd equation of motion:
[tex]2as_{2} = V_{f}^{2} - V_{i}^{2}\\\\[/tex]
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,
[tex]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[/tex]
So the total distance covered by the car before stopping is:
[tex]s = s_{1} + s_{2}\\s = 9.3\ m + 43.245\ m\\[/tex]
s = 52.545 m
