Constant Acceleration You are driving toward a traffic...

Constant Acceleration You are driving toward a traffic signal when it turns yellow. Your speed is the legal speed limit of $v_{0}=55 \mathrm{km} / \mathrm{h}$ ; your best deceleration rate has the magnitude $a=5.18 \mathrm{m} / \mathrm{s}^{2} .$ Your best reaction time to begin braking is $T=0.75$ s. To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at 55 $\mathrm{km} / \mathrm{h}$ if the distance to the intersection and the duration of the yellow light are (a) 40 $\mathrm{m}$ and 2.8 $\mathrm{s}$ and $(\mathrm{b}) 32 \mathrm{m}$ and 1.8 $\mathrm{s}$ ? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate).

Constant Acceleration
You are driving toward a traffic signal when it turns yellow. Your speed is the legal speed limit of $v_{0}=55 \mathrm{km} / \mathrm{h}$ ; your best deceleration rate has the magnitude $a=5.18 \mathrm{m} / \mathrm{s}^{2} .$ Your best reaction time to begin braking is $T=0.75$ s. To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at 55 $\mathrm{km} / \mathrm{h}$ if the distance to the intersection and the duration of the yellow light are (a) 40 $\mathrm{m}$ and 2.8 $\mathrm{s}$ and $(\mathrm{b}) 32 \mathrm{m}$ and 1.8 $\mathrm{s}$ ? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate).

Key Concepts

Kinematic Equations for Constant Acceleration

This concept involves using the equations of motion to analyze objects moving under a constant acceleration. In problems involving deceleration or acceleration, these equations (such as d = v?t + ½at² and v² = v?² + 2ad) allow the calculation of distances, times, and final velocities, which are critical in determining whether an object can safely come to a stop or cover a certain distance within a specified time.

Reaction Time in Motion Analysis

Reaction time is the delay between the perception of a stimulus and the initiation of a response, such as applying brakes. In motion problems, the distance the vehicle travels during this reaction time must be added to other distance calculations like braking distance. This ensures that the total distance required to safely stop or maneuver is correctly accounted for.

Deceleration as Negative Acceleration

Deceleration is the process of reducing velocity and is treated mathematically as negative acceleration. Understanding deceleration is essential in determining how quickly an object can come to a stop, as it directly influences the calculation of stopping distance and the time required to safely reduce speed.

Comparative Analysis of Stopping Distance versus Available Distance

This concept involves comparing the total distance required to safely stop (including both the distance traveled during the reaction period and the braking distance) against the available distance to an obstacle or intersection. It is crucial for making informed decisions about whether to continue at current speed or to initiate braking to avoid unsafe situations.

Transcript

00:01 So here we know that for the, we can read our given, so our acceleration is negative 5 .18 meters per second square.

00:09 This is our fastest deceleration.

00:12 We know that the initial, we would say this is going to be equal to 55 kilometers per hour, multiplied by 1 ,000 meters per kilometer, multiplied by one hour for every 3 ,600 seconds.

00:28 And this is equaling approximately 15 .28 meters per second.

00:35 We can say for part a, the velocity here is constant during the reaction time.

00:41 So the best reaction time that we have is 0 .75 seconds.

00:51 And so we can say that the distance traveled during our reaction time would simply be our initial velocity times t.

01:00 This would be equal to 15 .28 meters per second multiplied by 0 .75 seconds.

01:12 And this is equaling 11 .46 meters.

01:16 So before even starting to break, we have already traveled to 11 .46 meters.

01:21 Now we can then use equation 216.

01:26 This would yield that v final squared equal v initial squared plus 2a times.

01:34 The distance during breaking so essentially we can say that v final squared we know to be zero because we're trying to come to a stop and we have that the distance travel during breaking would be equal to negative the initial squared divided by 2a so the distance travel during breaking would be negative 15 .28 meters per second quantity squared this would be divided by two times negative 5 .18 meters per second squared and we have the distance during braking to be 22 .53 meters so take note of this take note of this we know that then at this then at this point the total for part again for part a the total distance would be equal to 22 .53 meters plus 34 .0 meters this is going to be equal to rather 22 .53 meters plus our reaction time of 11 .46 meters.

02:45 This is approximately equal to 34 .0 meters.

02:50 My apologies.

02:52 Now here if it's 34 point meters here the driver can stop in time.

02:58 So driver will be able to stop in time.

03:09 Now if the driver were to continue driving at the speed limit, we can say that t would be equal to 40 meters divided by 15 .28 meters per second.

03:27 This is equaling 2 .6 seconds.

03:31 This is essentially barely enough time to enter the intersection before the light turns.

03:40 So here driver can enter intersection before light turns red.

03:56 And usually if you enter the intersection before the light turns red, usually people, i don't, of course i don't know the laws, but usually people would find that acceptable.

04:09 So we could say either method works...

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