Finding the distance, once you have acceleration and time

In summary, an alcohol free driver needs 0.8 seconds to stop after seeing an emergency. A person who has had 4 beers needs 2 seconds, a person who has had five, needs 3 seconds. If driving at speed of 17 m/s, the reaction distance is 31.25 meters for a person who has had 4 beers, 41.25 meters for a person who has had 5 beers, and 51.875 meters for a person who has had 6 beers.
  • #1
christinaa_s
15
0
QUESTION: An alcohol free driver needs 0.8 s to stop after seeing an emergency. A person who has had 4 beers need 2 seconds, a person who has had five, need 3 seconds. If driving at speed of 17 m/s, find the reaction distance (after seeing emergency, the distance traveled before pressing brakes) for the all three times. What if the speed is 25 m/s. What if it is 33 m/s.

so first i had to find acceleration, then the reaction time.

for the non-alcoholic the acceleration is:
for 25m/s acceleration = 31.25
for 33 m/s acceleration = 41.25

if i have the acceleration how do i find the distance it took to stop. And is this acceleration supposed to be negative?
 
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  • #2
To find acceleration, I take it you used:

[tex]v_f = v_i + at[/tex]
(acceleration is negative)

To find position, the equation is very similar:

[tex]s_f = s_i + v_i t + 1/2 a t^2[/tex]

In fact, the first equation is the derivative with respect to time of the second equation.
 
  • #3
christinaa_s said:
QUESTION: An alcohol free driver needs 0.8 s to stop after seeing an emergency. A person who has had 4 beers need 2 seconds, a person who has had five, need 3 seconds. If driving at speed of 17 m/s, find the reaction distance (after seeing emergency, the distance traveled before pressing brakes) for the all three times. What if the speed is 25 m/s. What if it is 33 m/s.

so first i had to find acceleration, then the reaction time.

for the non-alcoholic the acceleration is:
for 25m/s acceleration = 31.25
for 33 m/s acceleration = 41.25

if i have the acceleration how do i find the distance it took to stop. And is this acceleration supposed to be negative?


The question does not ask for the distance to come to a complete stop! It asks "the distance traveled before pressing brakes". There is no acceleration in this problem. They just want the distance traveled before the person starts braking (and during that time, the speed is constant so it's easy!)

The accelerations you calculated do not make sense. It is as if you are saying "for a alcohol free driver, it wil take 0.8 second to come to rest". This is NOT the meaning of the 0.8 second!

Patrick
 
  • #4
BobG said:
(acceleration is negative)

I hope you mean deceleration :wink:
 
  • #5
Acceleration can perfectly well be negative.

We use the term deceleration only if we want to emphazise that the SPEED is being reduced.

When that is not a concern, the term "deceleration" is inappropriate.
 
  • #6
arildno said:
Acceleration can perfectly well be negative.

We use the term deceleration only if we want to emphazise that the SPEED is being reduced.

When that is not a concern, the term "deceleration" is inappropriate.

hear hear!

The danger is that students make the wrong assumption that deceleration implies negative acceleration and acceleration implies positive acceleration.

In everyday language, deceleration means that the speed decreases, as pointed out by Arildno. But this may occur for [itex] a_x >0 [/itex] or for [itex] a_x <0 [/itex] depending on the sign of [itex] v_x[/itex]. It is quite dangerous to use the term deceleration with students, in my experience. I prefer to stick with "positive acceleration" or "negative acceleration".

I like to talk about whether an object speeds up or slows down, but I make it clear to my student that each case may occur with a either positive or negative a_x. The rule being that if both a_x and v_x have the same sign, the object speeds up. If they have opposite signs, the object slows down.

Pat
 
  • #7
Good point nrged, now I know where all my mistakes come from in kinematics:redface:
 
  • #8
nrqed said:
The question does not ask for the distance to come to a complete stop! It asks "the distance traveled before pressing brakes". There is no acceleration in this problem. They just want the distance traveled before the person starts braking (and during that time, the speed is constant so it's easy!)

The accelerations you calculated do not make sense. It is as if you are saying "for a alcohol free driver, it wil take 0.8 second to come to rest". This is NOT the meaning of the 0.8 second!

Patrick

Bravo! So stupid of me to miss something like that :mad: Interestingly worded problem, though :)
 
  • #9
Thank-you so much for your help!

Nrged, you are completely right, I misunderstood the question. It is asking for the "reaction distance" so the distance driven before the brakes were applied, which is at a constant speed. So the equation I use is d = v * t .

Also, how do I draw a vector diagram to scale?
 
  • #10
QUESTION: An alcohol free driver needs 0.8 s to stop after seeing an emergency. A person who has had 4 beers need 2 seconds, a person who has had five, need 3 seconds. If driving at speed of 17 m/s, find the reaction distance (after seeing emergency, the distance traveled before pressing brakes) for the all three times. What if the speed is 25 m/s. What if it is 33 m/s.

Cannot be solved: All we know is the initial velocity and how long it took to stop after seeing the emergency. The question asks how long the driver took to react from seeing the accident to pressing the brakes. To solve this we must know the affectiveness of the brakes to determine what part of the stopping time was used by braking and what part was used by getting to the brakes or reacting. The answer will be different depending the affectiveness of the brakes. If when the brakes are pressed it takes .1 sec to stop the car at 17m/s then the reaction time for the alcohol free driver is .7 sec.

reaction time = total stopping time - braking time.
 
  • #11
JHWilliams said:
Cannot be solved: All we know is the initial velocity and how long it took to stop after seeing the emergency. The question asks how long the driver took to react from seeing the accident to pressing the brakes. To solve this we must know the affectiveness of the brakes to determine what part of the stopping time was used by braking and what part was used by getting to the brakes or reacting. The answer will be different depending the affectiveness of the brakes. If when the brakes are pressed it takes .1 sec to stop the car at 17m/s then the reaction time for the alcohol free driver is .7 sec.

reaction time = total stopping time - braking time.
That was my thought as well. I suspect the wording of the question doesn't match the intent of the question, but if it's taken literally as written, I don't see how it can be solved.

Since the actual stopping time would vary with speed, but there's no mention of which speed the given times apply to, I suspect that those times are reaction times rather than stopping times.
 

1. How do you calculate distance when given acceleration and time?

The formula for calculating distance when given acceleration and time is: d = 1/2at^2, where d is the distance, a is the acceleration, and t is the time.

2. Can you use the same formula to find distance for any type of motion?

Yes, the formula for calculating distance when given acceleration and time can be used for any type of motion, as long as the acceleration remains constant throughout the entire duration of the motion.

3. What units should be used for acceleration and time in the formula?

Acceleration should be in meters per second squared (m/s^2) and time should be in seconds (s) in order for the formula to yield the distance in meters (m).

4. Is it necessary to convert any units before plugging them into the formula?

Yes, it is important to ensure that all units are consistent and in the correct form before using them in the formula. If necessary, conversions should be made before plugging the values into the formula.

5. Can this formula be used for finding distance in non-uniform motion?

No, this formula is only applicable for calculating distance in uniform motion, where the acceleration remains constant. For non-uniform motion, a different formula must be used to find the distance.

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