A driver traveling at 50mph is 80m from a wall ahead. If the driver applies the brake
immediately at a reaction time of 2 seconds and begins slowing the vehicle at 10m/s2.
a. Find the distance from the stopping point to the wall.
b. Determine the braking time or time during deceleration
a. Well, let's see here. The driver's reaction time is 2 seconds, so during that time the car keeps moving at a constant speed of 50mph. Now, 50mph is about 80.5 km/h, which is around 22.35 m/s. So in those 2 seconds, the car travels a distance of 2 * 22.35 = 44.7 meters.
After the reaction time, the driver starts slowing down the car at 10m/s^2. Now, we need to find out the distance the car travels during deceleration until it comes to a stop. We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
For deceleration, the final velocity is 0 (because the car comes to a stop), the initial velocity is 22.35 m/s, and the acceleration is -10m/s^2 (negative because it's deceleration). Plugging these values in, we have:
0^2 = 22.35^2 + 2(-10)s
0 = 499.8225 - 20s
20s = 499.8225
s = 24.99 meters
So the distance from the stopping point to the wall is 44.7 + 24.99 = 69.69 meters. Quite the distance!
b. To determine the braking time or time during deceleration, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time elapsed.
Again, for deceleration, the final velocity is 0, the initial velocity is 22.35 m/s, and the acceleration is -10m/s^2. Plugging these values in, we have:
0 = 22.35 + (-10)t
-22.35 = -10t
t = 2.235 seconds
So the braking time or time during deceleration is 2.235 seconds. That's the amount of time it takes for the car to come to a stop. Let's hope the driver doesn't spill their coffee during those 2 seconds!
To solve this problem, we can use the equations of motion to find the distance from the stopping point to the wall and the braking time.
a. To find the distance from the stopping point to the wall, we need to consider the distance the car travels during the driver's reaction time and the distance it travels during deceleration.
During the driver's reaction time of 2 seconds, the car continues to move at its initial velocity of 50 mph (which we'll convert to m/s).
Let's convert the initial velocity from mph to m/s:
50 mph = (50 * 1609.34) / 3600 = 22.35 m/s (approx)
Distance traveled during reaction time = initial velocity * reaction time
= 22.35 m/s * 2 s
= 44.7 m
After the reaction time, the car starts decelerating at a rate of 10 m/s^2. We can use the equation of motion:
Final velocity (v) = Initial velocity (u) + (acceleration (a) * time (t))
The final velocity when the car comes to a stop is 0 m/s. Let's assume the time for deceleration is t seconds.
Using the equation, we can rewrite it to solve for time:
Final velocity (v) = Initial velocity (u) + (acceleration (a) * time (t))
0 = 22.35 m/s + (-10 m/s^2) * t
Rearranging the equation:
10t = 22.35 m/s
Time, t = 22.35 m/s / 10 m/s^2
t = 2.235 s
The total distance traveled during deceleration can be calculated using the equation:
Distance traveled during deceleration = (Initial velocity * time) + (0.5 * acceleration * time^2)
= (22.35 m/s * 2.235 s) + (0.5 * -10 m/s^2 * (2.235 s)^2)
= 50 m + (-25 m)
= 25 m
Therefore, the total distance from the stopping point to the wall is the sum of the distance traveled during the driver's reaction time and the distance traveled during deceleration:
Total distance = distance during reaction time + distance during deceleration
= 44.7 m + 25 m
= 69.7 m (approx)
b. The braking time, or time during deceleration, is found to be 2.235 seconds. This is the time it takes for the car to come to a complete stop after the driver applies the brakes.
To solve this problem, let's break it down step by step.
a. Find the distance from the stopping point to the wall.
First, let's calculate the distance covered during the driver's reaction time. We can do this by using the formula:
Distance = Initial velocity * Time
Given that the initial velocity is 50 mph and the reaction time is 2 seconds, we need to convert the velocity from mph to m/s to have consistent units:
Initial velocity = 50 mph = (50 * 1609.34) / 3600 m/s ā 22.35 m/s
Now, we can calculate the distance covered during the reaction time:
Distance = 22.35 m/s * 2 s = 44.7 m
Next, let's calculate the distance covered during the deceleration phase using the formula:
Distance = Initial velocity * Time + 0.5 * Acceleration * Time^2
Given that the initial velocity is 22.35 m/s and the deceleration is -10 m/s^2 (negative because it is a deceleration), we need to find the time it takes to stop, denoted as T.
Rearranging the formula, we get:
0 = 22.35 m/s * T + 0.5 * (-10 m/s^2) * T^2
Simplifying this quadratic equation by dividing the entire equation by T:
0 = 22.35 - 5T
5T = 22.35
T = 4.47 s
Now, let's substitute this value back into the distance formula to find the distance covered during the deceleration phase:
Distance = 22.35 m/s * 4.47 s + 0.5 * (-10 m/s^2) * (4.47 s)^2
Distance = 99.8 m + 0.5 * (-10 m/s^2) * 19.9809 s^2
Distance ā 99.8 - 99.9 m
Finally, to find the total distance from the stopping point to the wall, we sum the distances covered during the reaction time and the deceleration phase:
Total distance = Distance during reaction time + Distance during deceleration
Total distance = 44.7 m + (-0.1) m ā 44.6 m
Therefore, the distance from the stopping point to the wall is approximately 44.6 meters.
b. Determine the braking time or time during deceleration
In the previous calculation, we found that the time taken to stop during the deceleration phase is approximately 4.47 seconds. Therefore, the braking time is 4.47 seconds.
Vo = 50mi/1h * 1h/3600s * 1600m/mi = 22.2 m/s.
d = 80-22.2*2 = 35.6 m. = distance from wall after 2-second reaction time.
a. V^2 = Vo^2+2a*d = 0
22.2^2-20d = 0
d1 = 24.6 m. = stopping point.
d2 = 35.6-24.6 = 11 m. from wall.
b. V = Vo+a*T = 0
22.2-10T = 0
T =