A driver of a car going 95 km/h suddenly sees the lights of a barrier ahead. It takes the driver 0.55 s before he applies the brakes (this is known as reaction time). Once he does begin to brake, he accelerates at a rate of -10.0 m/s²
a.) How far did the driver travel during reaction time?
b.) How far has the driver traveled once the car comes to a complete stop?
c.) f the barrier is 40 meters away , does the driver have enough time to come to a complete stop?
95 km/hr = 26.39 m/s
a) s = vt = 26.39 m/s * 0.55s = 14.51 m
b) 26.39/10 = 2.64 s stopping time, so
s = 14.51 + 26.29 * 2.64 - 5*2.64^2 = 49.07 m
c) nope
well, technically, yes, since he will come to an abrupt stop at the barrier!
thanks!
To find the answers to these questions, we need to use the equations of motion.
a.) Firstly, let's calculate the distance the driver traveled during the reaction time. We know that the initial velocity (u) of the car is 95 km/h, which can be converted to meters per second (m/s) by dividing by 3.6:
u = 95 km/h = (95 * 1000) / 3600 = 26.39 m/s
Using the formula for distance traveled during constant acceleration:
distance = (initial velocity * time) + (0.5 * acceleration * time²)
Here, the initial velocity (u) is 26.39 m/s, acceleration (a) is 0 m/s² (because the driver has not yet applied the brakes), and the time (t) is 0.55 seconds:
distance = (26.39 * 0.55) + (0.5 * 0 * (0.55)²
= 14.52 meters
Therefore, the driver has traveled 14.52 meters during the reaction time.
b.) Now, let's calculate the distance the driver traveled once the car comes to a complete stop. We can use the same formula as above, but this time the acceleration is -10.0 m/s²:
distance = (0 * time) + (0.5 * (-10.0) * time²)
Since the final velocity (v) is 0 m/s when the car comes to a complete stop, we can use the third equation of motion:
v² = u² + 2as
Here, the initial velocity (u) is 0 m/s, acceleration (a) is -10.0 m/s², and the distance (s) is what we need to find:
0 = 0² + 2 * (-10.0)s
Simplifying the equation:
-20s = 0
Therefore, the distance (s) is 0 meters. This means that the car comes to a stop at the same point where it started braking.
c.) The barrier is 40 meters away. Since the total distance traveled during the reaction time and when coming to a complete stop is 14.52 meters, the total distance needed to stop is:
40 meters - 14.52 meters = 25.48 meters
To determine if the driver has enough time to stop, we can now calculate the time it takes to cover the remaining distance.
Using the third equation of motion again:
v² = u² + 2as
Here, the initial velocity (u) is 26.39 m/s (the speed when the driver notices the barrier), the acceleration (a) is -10.0 m/s², and the distance (s) is the remaining distance of 25.48 meters:
0 = (26.39)² + 2 * (-10.0) * s
Solving for s:
(26.39)² = 2 * (-10.0) * s
s = ((26.39)²) / (-20)
s ≈ 34.84 meters
Since the remaining distance is 25.48 meters, which is less than the 34.84 meters needed to stop, the driver does not have enough time to come to a complete stop before reaching the barrier.