How far does the car travel before its brakes are applied to slow down?

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Homework Help Overview

The discussion revolves around the motion of two cars, specifically focusing on the blue car's acceleration, distance traveled before braking, and the yellow car's acceleration. The blue car accelerates uniformly, travels at a constant speed, and then decelerates to a stop, while the yellow car accelerates uniformly over the same distance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the motion of the blue car in segments: initial acceleration, constant speed, and deceleration. They question the calculations for distance and acceleration, and some participants suggest breaking down the problem into smaller steps.

Discussion Status

Participants have provided various calculations and attempted to clarify the relationships between the blue and yellow cars. There is ongoing exploration of the blue car's distance before braking and the yellow car's acceleration, with some guidance offered on using kinematic equations.

Contextual Notes

Participants note the total distance and time for both cars, but express difficulty in relating the different acceleration profiles of the two vehicles. The discussion includes questioning assumptions about uniform acceleration and the implications for the yellow car's motion.

ctwokay
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Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 5 m/s2 for 4.3 seconds. It then continues at a constant speed for 9.7 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 286 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop


1. How far does the blue car travel before its brakes are applied to slow down?
2. What is the acceleration of the blue car once the brakes are applied?
3. What is the total time the blue car is moving?
4. What is the acceleration of the yellow car?

1. I tried using 14s to calculate but to no avail.
 
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Start by figuring out the motion of the blue car during each segment: acceleration 1, constant speed, acceleration 2. Find the time and distance of each segment.
 
i calculate for 14s is the car travel before it slows down is equal to 59.25 is that correct?
 
ctwokay said:
i calculate for 14s is the car travel before it slows down is equal to 59.25 is that correct?
No.

Show how you did the calculation.

Or just do it step by step:
How far does the blue car travel during the first 4.3 seconds?
What is its speed at the end of the first 4.3 seconds?
How far does it travel during the next 9.7 seconds?
 
Doc Al said:
No.

Show how you did the calculation.

Or just do it step by step:
How far does the blue car travel during the first 4.3 seconds?
What is its speed at the end of the first 4.3 seconds?
How far does it travel during the next 9.7 seconds?

it travel 1/2*5*4.3=10.25m
its speed is 5*4.3=21.5m
for the next 9.7s travel 9.7*5=48.5m
 
ctwokay said:
it travel 1/2*5*4.3=10.25m
No. The formula should be: x = ½at2.
its speed is 5*4.3=21.5m
Right. 21.5 m/s.
for the next 9.7s travel 9.7*5=48.5m
No. You just said that the speed was 21.5 m/s, so use that fact to calculate the distance.
 
Doc Al said:
No. The formula should be: x = ½at2.

Right. 21.5 m/s.

No. You just said that the speed was 21.5 m/s, so use that fact to calculate the distance.

ok thank you i got the answer is 254.775m
 
ctwokay said:
ok thank you i got the answer is 254.775m
Good! Keep going.
 
2. the acceleration i use d=286-254.775=31.225m
t=31.225/21.5=1.45s
then a=21.5/1.45=14.8m/s^2
is that correct?
 
  • #10
ctwokay said:
2. the acceleration i use d=286-254.775=31.225m
OK.
t=31.225/21.5=1.45s
No, since the speed is not constant. What's the average speed during that segment of the motion?
 
  • #11
Doc Al said:
No, since the speed is not constant. What's the average speed during that segment of the motion?

ave speed = (0-21.5)/2=-10.75m/s
 
  • #12
i got the answer i use d=31.225,t=31.225/10.75=2.9s
31.225=1/2at^2
a=-7.4m/s^2
 
  • #13
acceleration for the yellow car is it related to the blue car?
 
  • #14
ctwokay said:
ave speed = (0-21.5)/2=-10.75m/s
Good.
ctwokay said:
i got the answer i use d=31.225,t=31.225/10.75=2.9s
31.225=1/2at^2
a=-7.4m/s^2
Good.

You can also use a = Δv/Δt to get the same answer.
 
  • #15
ctwokay said:
acceleration for the yellow car is it related to the blue car?
Sure. First figure out the answer to part 3.
 
  • #16
Doc Al said:
Sure. First figure out the answer to part 3.

total time is 17s
 
  • #17
i can't seem to relate the yellow to the blue car.
Both have same distance and same time but different acceleration.
 
  • #18
ctwokay said:
total time is 17s
Be more accurate than that.

Use that time and the total distance to figure out the acceleration that the yellow car must have.
 
  • #19
ctwokay said:
i can't seem to relate the yellow to the blue car.
Both have same distance and same time but different acceleration.
Sure, the overall distance and time is the same, but the motions are totally different. The blue car accelerates for a bit, then travels at constant speed for a bit, then accelerates again to slow down. But the yellow car simply accelerates at one rate for the whole distance. And at the end of that time and distance the blue car is at rest but the yellow car is not.
 
  • #20
if i use a=v/t for yellow car,i have to find the v of yellow car then i use the value of v to find a,is that correct?
But when i get the value for yellow it does not seem right.
 
  • #21
ctwokay said:
if i use a=v/t for yellow car,i have to find the v of yellow car then i use the value of v to find a,is that correct?
But when i get the value for yellow it does not seem right.
You have the distance and time, so use a kinematic formula (which you've already used in this thread) to solve for the acceleration.
 

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