A person driving her car at 56km/hr approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 20.s before turning to red, and she is 30m away from the near side of the intersection. The intersection is 15m wide. Her car's maximum deceleration is -6.2m/s^2, whereas it can accelerate from 56km/hr to 70km/hr in 4.2s. Ignore the length of her car and her reaction time.
a.) if she hits the brakes, how far will she travel before stopping?
b.) if she hits the gas instead, how far will she travel before the light turns red?
c.) should she try to stop or should she speed up to cross the intersection before the light turns red?
use
x = x0 + v0*t + 1/2*a*t^2
v = v0 + a*t
where x is position, x0 is initial position, v0 is initial speed, a is acceleration/deceleration, t is time, and v is speed as a function of time
a) 56 - 6.2*t^2 = 0
Solve for t, the time it takes for the speed to go to zero. The distance traveled in this time is then 1/2*6.2*t^2
b) 56 + a*4.2 = 70
Solve for a, the acceleration
Then calculated the distance she travels (I would assume that she stops accelerating after she hits 70 km/hr
So for the first 4.2 seconds, she travels
x1 = 56*t + 1/2*a*4.2^2
and then she travels an additional distance:
x2 = 70*(20-4.2)
for a total distance of x1 + x2
c) Evaluate where she will be in part a) and b) with respect to the traffic lights and decide what she should do
To solve this problem, we can use the kinematic equations of motion to determine the distance the person will travel before stopping or reaching the intersection. Let's calculate each scenario step by step:
a) If she hits the brakes, we need to calculate the distance traveled during the deceleration phase. The equation we need to use is:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (0 m/s in this case, since she is stopping)
vi = initial velocity (56 km/hr converted to m/s)
a = acceleration (-6.2 m/s^2)
d = distance traveled
Converting the initial velocity to m/s, we have:
vi = 56 km/hr * 1000 m/3600 s = 15.6 m/s
Plugging in the values into the equation, we can solve for d:
0 = (15.6 m/s)^2 + 2 * (-6.2 m/s^2) * d
Rearranging the equation, we get:
2 * (6.2 m/s^2) * d = (15.6 m/s)^2
Now, solve for d:
d = (15.6^2 m^2/s^2) / (2 * 6.2 m/s^2)
d ≈ 19.62 m
Therefore, if she hits the brakes, she will travel approximately 19.62 meters before stopping.
b) If she hits the gas and accelerates instead, we need to calculate the distance traveled during the acceleration phase. The equation we need to use is:
vf = vi + at
Where:
vf = final velocity (70 km/hr converted to m/s)
vi = initial velocity (56 km/hr converted to m/s)
a = acceleration (to be calculated)
t = time taken to accelerate (4.2 s)
Converting the velocities to m/s, we have:
vi = 56 km/hr * 1000 m/3600 s = 15.6 m/s
vf = 70 km/hr * 1000 m/3600 s = 19.4 m/s
Plugging in the values into the equation, we can solve for a:
19.4 m/s = 15.6 m/s + a * 4.2 s
Rearranging the equation, we get:
a * 4.2 s = 19.4 m/s - 15.6 m/s
Now, solve for a:
a = (19.4 m/s - 15.6 m/s) / 4.2 s
a ≈ 0.905 m/s^2
Now, to find the distance traveled during this acceleration phase, we'll use the formula:
d = vi * t + 1/2 * a * t^2
Plugging in the values:
d = 15.6 m/s * 4.2 s + 1/2 * 0.905 m/s^2 * (4.2 s)^2
d ≈ 71.46 m
Therefore, if she hits the gas, she will travel approximately 71.46 meters before the light turns red.
c) To determine whether she should try to stop or speed up, we need to compare the distances traveled in both scenarios to the distance remaining before the intersection.
For the braking scenario, she would travel approximately 19.62 m, which is less than the distance remaining before the near side of the intersection, which is 30 m. Therefore, she should try to stop.
In the acceleration scenario, she would travel approximately 71.46 m, which is more than the distance remaining before the near side of the intersection. Therefore, she should not try to speed up to cross the intersection before the light turns red.
In conclusion, she should hit the brakes and try to stop before reaching the intersection.
To solve this problem, we'll need to analyze the motion of the car under two scenarios: (a) when the driver hits the brakes and (b) when the driver hits the gas. Let's break it down step by step.
a) If the driver hits the brakes, we need to find out how far the car will travel before coming to a stop. We can use the equation of motion:
vf^2 = vi^2 + 2ad
where vf is the final velocity (0 m/s in this case, as the car stops), vi is the initial velocity (56 km/hr), a is the deceleration (-6.2 m/s^2), and d is the distance traveled.
First, let's convert the initial velocity from km/hr to m/s:
vi = 56 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 15.6 m/s
Using the equation of motion, we can find the distance traveled:
0 = (15.6 m/s)^2 + 2*(-6.2 m/s^2) * d
Rearranging the equation, we get:
15.6^2 = 2*(-6.2) * d
Simplifying the equation, we find:
d = (15.6^2) / (-2*6.2) = 15.6 m^2 / 6.2 = 25.2 m
Therefore, if the driver hits the brakes, she will travel approximately 25.2 meters before stopping.
b) If the driver hits the gas, we need to find out how far the car can travel before the light turns red. Since we're given the car's acceleration from 56 km/hr to 70 km/hr in 4.2 s, we can use the average acceleration formula:
a_avg = Δv / Δt
where Δv is the change in velocity (70 km/hr - 56 km/hr) and Δt is the time interval (4.2 s). We can then use this average acceleration to calculate the distance traveled using the equation of motion.
First, let's convert the velocity change and interval to m/s:
Δv = (70 km/hr - 56 km/hr) * (1000 m/1 km) * (1 hr/3600 s) = 3.9 m/s
Δt = 4.2 s
Using the average acceleration formula, we find:
a_avg = 3.9 m/s / 4.2 s ≈ 0.93 m/s^2
Now, we can determine the distance traveled using the equation of motion:
vf^2 = vi^2 + 2ad
Here, vf is the final velocity (70 km/hr), vi is the initial velocity (56 km/hr), a is the average acceleration (0.93 m/s^2), and d is the distance traveled.
First, convert the initial and final velocities to m/s:
vi = 56 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 15.6 m/s
vf = 70 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 19.4 m/s
Using the equation of motion, we can find the distance traveled:
(19.4 m/s)^2 = (15.6 m/s)^2 + 2*(0.93 m/s^2) * d
Simplifying the equation, we find:
d = [(19.4 m/s)^2 - (15.6 m/s)^2] / (2*0.93 m/s^2) ≈ 118.1 m
Therefore, if the driver hits the gas, she will travel approximately 118.1 meters before the light turns red.
c) To determine whether the driver should try to stop or speed up, we need to compare the distances traveled in both scenarios.
In scenario (a), the driver would travel approximately 25.2 meters before stopping.
In scenario (b), the driver would travel approximately 118.1 meters before the light turns red.
Since 25.2 meters is significantly shorter than 118.1 meters, the driver should try to stop the car rather than trying to speed up. This way, she can safely come to a stop before reaching the intersection.
It's worth noting that this solution assumes ideal conditions and neglects factors like the length of the car and the driver's reaction time, as stated in the problem. In real-life scenarios, these factors should be taken into account for accurate decision-making.