An inattentive driver is travelling 18m/s. When he notice a red light ahead. His car is capable of a breaking acceleration of 3.65m/s^2. If it takes him 0.200 second to get the brakes on and he is 45m from the intersection when he sees the ligth, will he be able to stop in time?
V^2 = Vo^2 + 2a*d = 0.
324 + (-7.30)d = 0,
d = 44.4 m = stopping distance.
45 - 18m/s * 0.2s = 41.4 m. = Required stopping distance.
No, he will not be able to stop in time.
just plug in your numbers into the equation of motion:
how long does it take to stop (v=0)?
v = 18 - 3.65t
t = 4.932
How far does he go?
s = vt + 1/2 at^2
s = 18t - 1.825t^2
time to stop: 4.932+0.200 = 5.132 seconds
s = 18*5.132 - 1.825*5.132^2 = 44.310
>whew< close call!
Well, let's calculate and see if our inattentive driver will make it in time to stop! We can use the good old kinematic equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time taken.
Now, let's break it down. The driver's initial velocity is 18 m/s, the acceleration is -3.65 m/s^2 (negative because it's braking), and the time taken is 0.2 seconds.
Using the equation, we can find the final velocity after braking:
vf = 18 + (-3.65) * 0.2
vf = 18 + (-0.73)
vf = 17.27 m/s
Bu-bu-bu-but we're not done yet! We also need to know the distance traveled during this time.
To find that, we can use another fun equation:
d = vit + 0.5at^2
Where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time taken.
Now, let's plug in the values:
d = 18 * 0.2 + 0.5 * (-3.65) * (0.2)^2
d = 3.6 + 0.5 * (-3.65) * 0.04
d = 3.6 + 0.5 * (-0.146)
d = 3.6 - 0.073
d = 3.527 m
Oh dear! The driver's car would have traveled approximately 3.527 meters during the reaction time.
So, if our inattentive driver is 45 meters away from the intersection when he sees the red light, it seems like he won't be able to stop in time. Looks like he's going to have an unplanned close encounter with that red light! Time to take a deep breath and hope for the best.
To determine if the driver will be able to stop in time, we can use the equation of motion:
v² = u² + 2as
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- s is the distance
In this case, the driver is initially traveling at a velocity of 18 m/s, the acceleration during braking is -3.65 m/s² (negative because it is deceleration), and the distance to the intersection is 45 m.
First, we need to find the final velocity, which is 0 m/s since the driver needs to come to a stop:
0 = 18² + 2 * (-3.65) * s
Simplifying the equation:
0 = 324 - 7.3s
Rearranging the equation:
7.3s = 324
s = 324 / 7.3
s ≈ 44.4
The driver will be able to stop in time if the distance to the intersection is less than 44.4 meters. In this case, the distance is 45 meters, so the driver will not be able to stop in time.
To determine whether the driver will be able to stop in time, we need to calculate the distance it takes for the car to come to a complete stop and see if it's less than or equal to the distance remaining to the intersection.
Let's break it down step by step:
Step 1: Calculate the distance it takes for the car to come to a complete stop.
We can use the equation of motion:
vf^2 = vi^2 + 2ad
where vf is the final velocity (which should be 0 m/s since the car comes to a stop), vi is the initial velocity (18 m/s), a is the braking acceleration (-3.65 m/s^2), and d is the distance.
Rearranging the equation, we get:
d = (vf^2 - vi^2) / (2a)
Plugging in the values:
d = (0^2 - 18^2) / (2 * -3.65)
d = (-324) / (-7.3)
d = 44.38 m (rounded to two decimal places)
So the car needs a distance of approximately 44.38 m to stop completely.
Step 2: Determine if the remaining distance is less than or equal to the distance required to stop.
The remaining distance to the intersection is given as 45 m.
Comparing the remaining distance (45 m) with the distance required to stop (44.38 m):
45 m > 44.38 m
Since the remaining distance is greater than the distance required to stop, the driver will not be able to stop in time.
Therefore, the driver will not be able to stop before reaching the intersection.