A car is traveling at 50km/h. The driver sees a child run out into the road 5m ahead. She applies the breaks and the car stops in 5s. The driver's thinking time is 1.5s.
a. Will the car stop in time?
b. If the driver's thinking time is increased to 2.5s, will the car stop in time?
c. What happens if the thinking time is 1.5s but the car is traveling at 64km/h?
Are you sure the child is only 5 m ahead?
Vo = 50km/h = 50,000m/3600s = 13.89 m/s.
a. V = Vo + a*t = 0.
13.89 + a*5 = 0, a = -2.80 m/s^2.
V^2 = Vo^2 + 2a*d.
V = 0, Vo = 13.89 m/s, a = -2.80m/s^2, d = ?.
b.
Initial u=14m/s a=-2.8m/s^2 s=5 the time needed to stop can be find 5m=14t-2.8t^2÷2
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To answer these questions, let's break down the information given and apply some basic physics principles.
a) Will the car stop in time?
To determine if the car will stop in time, we need to consider the total distance the car will travel while the driver reacts to the child running out and while the car is decelerating to a stop.
The driver's thinking time is given as 1.5s, during which the car is still moving at 50km/h. To convert this speed to meters per second, we divide by 3.6 (since 1 km/h = 1/3.6 m/s). Therefore, the initial speed of the car is approximately 13.89 m/s.
During the thinking time, the car travels a distance equal to the product of the initial speed and the thinking time: (13.89 m/s) * (1.5 s) = 20.835 meters.
After the thinking time, the car starts decelerating to a stop. The braking time is given as 5s. During this time, the car goes from 50 km/h (or 13.89 m/s) to 0 m/s.
Using the equation for constant acceleration, 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, we can solve for the acceleration.
Rearranging the equation, a = (v^2 - u^2) / 2s, we find a = (0 - (13.89 m/s)^2) / (2 * s).
Plugging in the values, a = -191.875 m^2/s^2 / (2 * s).
To find the distance traveled during braking, we can use the equation: s = ut + (1/2)at^2. The initial velocity u is 13.89 m/s, the time t is 5 s, and the acceleration a is -191.875 m^2/s^2.
s = (13.89 m/s * 5 s) + (1/2) * (-191.875 m^2/s^2) * (5 s)^2.
Evaluating the equation, we find s = 69.45 meters.
Therefore, the total distance traveled during both the thinking and braking time is 20.835 meters (thinking time) + 69.45 meters (braking time) = 90.285 meters.
Since the child runs out into the road 5 meters ahead, the car will not stop in time and will collide with the child.
b) If the driver's thinking time is increased to 2.5s, will the car stop in time?
Using the same calculations as in part (a), we find that the total distance traveled during both the thinking and braking time is 20.835 meters (thinking time) + 69.45 meters (braking time) = 90.285 meters.
Since the child runs out 5 meters ahead, the car will stop in time to avoid the collision.
c) What happens if the thinking time is 1.5s, but the car is traveling at 64km/h?
To analyze this scenario, we can follow the same steps as in part (a) with revised initial velocity information.
Converting 64 km/h to m/s, we have 64 km/h ÷ 3.6 = 17.78 m/s.
The thinking time remains 1.5s, so during this time, the car travels a distance of (17.78 m/s) * (1.5 s) = 26.67 meters.
For the braking time, we still have 5s. Using the same acceleration equation as before, we have a = (0 - (17.78 m/s)^2) / (2 * s), where s is the distance.
Plugging in the values, we have a = -315.888 m^2/s^2 / (2 * s).
The distance traveled during braking is given by s = (17.78 m/s * 5 s) + (1/2) * (-315.888 m^2/s^2) * (5 s)^2.
Evaluating the equation, we find s = 222.225 meters.
The total distance traveled during both the thinking and braking time is 26.67 meters (thinking time) + 222.225 meters (braking time) = 248.895 meters.
Since the child runs out 5 meters ahead, the car will not stop in time and will collide with the child.
Therefore, in this scenario, even though the car is traveling at a higher speed, the longer thinking time does not compensate for the increased distance required to stop, resulting in a collision.