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

In summary: You should get 5 m/s2.You have the distance and time, so use a kinematic formula (which you've already used in this thread) to solve for the acceleration. You should get 5 m/s2.In summary, 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
  • #1
ctwokay
30
0
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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  • #2
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.
 
  • #3
i calculate for 14s is the car travel before it slows down is equal to 59.25 is that correct?
 
  • #4
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?
 
  • #5
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
 
  • #6
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.
 
  • #7
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
 
  • #8
ctwokay said:
ok thank you i got the answer is 254.775m
Good! Keep going.
 
  • #9
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.
 

1. How do you calculate the distance traveled by a car before its brakes are applied?

The distance traveled by a car before its brakes are applied can be calculated using the formula: distance = initial velocity * time + (1/2)(acceleration)(time)^2. This formula takes into account the initial velocity of the car, the time it takes for the brakes to be applied, and the acceleration of the car.

2. What factors affect the distance traveled by a car before its brakes are applied?

The distance traveled by a car before its brakes are applied can be affected by several factors, including the initial speed of the car, the reaction time of the driver, the condition of the brakes, and the surface of the road.

3. Can the distance traveled before brake application be reduced?

Yes, the distance traveled before brake application can be reduced by increasing the reaction time of the driver, maintaining the brakes in good condition, and choosing a road surface with better traction. Additionally, advanced braking technologies such as anti-lock braking systems (ABS) can also help reduce the distance traveled before brake application.

4. Is there a maximum distance that a car can travel before its brakes are applied?

No, there is no maximum distance that a car can travel before its brakes are applied. The distance traveled will depend on various factors, as mentioned earlier, and can vary greatly from one situation to another.

5. How can the distance traveled before brake application be measured?

The distance traveled before brake application can be measured using a speedometer, a stopwatch, and a known initial speed. The driver can note the time taken for the car to come to a complete stop after the brakes are applied, and the distance traveled can be calculated using the formula mentioned in the first question.

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