Reaction time while stopping a car

  • #1
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Homework Statement


To stop a car, first you require a certain reaction time to begin braking; then the car slows down under constant breaking decelaration. Suppose that the total distance moved by the car during these two phases is 56.7 m when its initial speed was 80.5 km/h and 24.4 m when its initial speed 48.3 km/h. What are
(a) The reaction time
(b) Magnitude of decelaration

2. Homework Equations

The Kinematical Equations

The Attempt at a Solution


Actually, my guess of what they're asking is the time between pressing the brake pedal and the brake pads squeezing in on the tyres.
So let to be the reaction time and a be the constant breaking decelaration.
u1 = 80.5 km/h = 22.36 m/s
u2 = 48.3 km/h = 13.42 m/s
Now, I thought of using s = ut + (1/2)at2, get two equations and solve them, but I have one basic question.
In the time interval I think they're asking, the car isn't decelarating. It starts decelarating after the brake pads squeeze in on the tyres. So how can I substitute to & a in the same equation?
 

Answers and Replies

  • #2
Doc Al
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Treat the "reaction time" period as having constant velocity. (The acceleration hasn't begun yet.) After the brakes are applied (and engaged), treat it as having constant acceleration.
 
  • #3
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56.7 = (22.36)(t + to) - (1/2)a(t2)
24.4 = (13.42)(t + to) - (1/2)a(t2)

Assuming t is the braking time and to is the reaction time.
3 variables, 2 equations...
 
  • #4
Doc Al
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56.7 = (22.36)(t + to) - (1/2)a(t2)
24.4 = (13.42)(t + to) - (1/2)a(t2)

3 variables, 2 equations...
Assuming t is the braking time and to is the reaction time.
Treat the motion as having two phases. For the accelerated phase, use a different kinematic formula that doesn't call for t.
 
  • #5
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For the accelerated phase, use a different kinematic formula that doesn't call for t.
There's only v2 = u2 - 2as. But v will be zero in each case. So I will have two inconsistent values of a.
 
  • #6
Ray Vickson
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Homework Statement


To stop a car, first you require a certain reaction time to begin braking; then the car slows down under constant breaking decelaration. Suppose that the total distance moved by the car during these two phases is 56.7 m when its initial speed was 80.5 km/h and 24.4 m when its initial speed 48.3 km/h. What are
(a) The reaction time
(b) Magnitude of decelaration

2. Homework Equations

The Kinematical Equations

The Attempt at a Solution


Actually, my guess of what they're asking is the time between pressing the brake pedal and the brake pads squeezing in on the tyres.
So let to be the reaction time and a be the constant breaking decelaration.
u1 = 80.5 km/h = 22.36 m/s
u2 = 48.3 km/h = 13.42 m/s
Now, I thought of using s = ut + (1/2)at2, get two equations and solve them, but I have one basic question.
In the time interval I think they're asking, the car isn't decelerating. It starts decelerating after the brake pads squeeze in on the tyres. So how can I substitute to & a in the same equation?

In this problem, "reaction time" is a sum: the time it takes for the foot to rise from the car floor to the brake pedal, plus the time needed to depress the pedal, plus the time you mentioned above. The assumption is that the total time for all of those effects is not affected by the initial speed, and that the deceleration itself is likewise independent of the initial speed. (That last one seems suspicious to me, as it assumes we push equally hard on the brake pedal whether we are slowing down in a controlled way before a red light or whether we are desperately trying to avoid a rear-end collision.)
 
  • #7
Doc Al
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There's only v2 = u2 - 2as. But v will be zero in each case. So I will have two inconsistent values of a.
The final velocity is zero, of course. But the initial velocity and distance will be different in each case.
 
  • #8
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(That last one seems suspicious to me, as it assumes we push equally hard on the brake pedal whether we are slowing down in a controlled way before a red light or whether we are desperately trying to avoid a rear-end collision.)
Yeah, that's quite ambiguous
 
  • #9
Doc Al
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@Ray Vickson is correct. But ignore those real-life details (:wink:) and solve the idealized problem: Treat the reaction time and acceleration as constants to be determined.
 
  • #10
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The final velocity is zero, of course. But the initial velocity and distance will be different in each case.
So, you mean a formula with distance, decelaration and initial velocity that doesn't include braking time but includes the reaction time? Then, I don't get what you're trying to say?
 
  • #11
Doc Al
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So, you mean a formula with distance, decelaration and initial velocity that doesn't include braking time but includes the reaction time? Then, I don't get what you're trying to say?
Don't try to jam it into one equation. (At least not at first.)

Phase 1: Constant speed for reaction time (Distance = D1)
Phase 2: You have the correct formula, which has distance, acceleration, and velocity (Distance = D2)

(You are given the total distance, D1 + D2.)
 
  • #12
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You are given the total distance, D1 + D2.
The equations in #3 do incorporate D1 + D2. That's why there's t + to in the first term but only t2 in the second.
By the way, if you subtract Eq. 1 and 2, you can obtain total time T, which comes out to be 3.61 s. Maybe we could use that.
 
  • #13
Doc Al
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The equations in #3 do incorporate D1 + D2. That's why there's t + to in the first term but only t2 in the second
I would skip those equations and use one that doesn't require t, which is an unknown. Like you had in #5.
 
  • #14
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I'm thinking of substitutimg t = u/a in Eq. 1 & 2, but that t + to term complicates it.
 
  • #15
Doc Al
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I'm thinking of substitutimg t = u/a in Eq. 1 & 2, but that t + to term complicates it.
Once again, why not just use the equation relating distance, acceleration, and speed? Then you won't have to go through the pain of eliminating the unnecessary time unknown.
 
  • #16
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But, that won't render anything... v is zero in both cases. I'll get 2 inconsistent values of a
 
  • #17
Doc Al
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But, that won't render anything... v is zero in both cases. I'll get 2 inconsistent values of a
The final velocity is zero! So what? You still have the initial velocity!
 
  • #18
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The final velocity is zero! So what? You still have the initial velocity!
u12 = 2as1
u22 = 2as2
If you substitute the given values, you get 2 different values of a.
 
  • #19
Orodruin
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##0 = u^2 - 2as## is valid only in the case of constant acceleration. This is only one of the phases of each of the situations. You are completely missing the reaction phase by just plugging the values into this equation.
 
  • #20
Doc Al
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u12 = 2as1
u22 = 2as2
If you substitute the given values, you get 2 different values of a.
That equation is only for the 2nd phase of the motion. Don't neglect the first phase. (The distance in that equation is not the total distance that was given in the problem statement.)
 
  • #21
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I'm still stuck at t + to. If only we could substitute t= u/a in t2 and that t in t + to would just magically vanish...
 
  • #22
Orodruin
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Things do not happen by magic in physics, they happen by logical reasoning given the assumed situation. You need to express the ##s## that should go into the equations in terms of the reaction time ##t_0## and the initial velocity ##v_i## and the total stopping distance ##s_0##. Once you have done that, you will have two equations with two unknowns (##t_0## and ##a##) which is a perfectly solvable system.
 
  • #23
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##0 = u^2 - 2as## is valid only in the case of constant acceleration. This is only one of the phases of each of the situations. You are completely missing the reaction phase by just plugging the values into this equation.
You need to express the ##s## that should go into the equations in terms of the reaction time ##t_0## and the initial velocity ##v_i## and the total stopping distance ##s_0##. Once you have done that, you will have two equations with two unknowns (##t_0## and ##a##) which is a perfectly solvable system.
That equation is only for the 2nd phase of the motion. Don't neglect the first phase. (The distance in that equation is not the total distance that was given in the problem statement.)
Yes, I get what you're saying mow. I was being an ass...
56.7 m = d11 + d12
= u1to + u12/2a
24.4 m = d21 + d22
= u2to + u22/2a
 
  • #24
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I'm getting to = 0.74 s and a = 6.23 m/s2.
 
  • #25
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Thank you all. And apologies for behaving rather idiotically
 
  • #26
verty
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I'm getting to = 0.74 s and a = 6.23 m/s2.

This is not quite right. You need to be more precise with the numbers.
 
  • #27
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This is not quite right. You need to be more precise with the numbers.
2-decimal accuracy is more than enough...
Are you getting a different answer?
The values don't satisfy the equation?
 
  • #28
Orodruin
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2-decimal accuracy is more than enough...
That is not what he is saying, he is saying you have a rounding error.
 
  • #29
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That is not what he is saying, he is saying you have a rounding error.
I did them using a calculator
a = 6.23724686 m/s2
to = 0.74098462 s
6.23 or 6.24 hardly matters in my case.
 
  • #30
haruspex
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I did them using a calculator
a = 6.23724686 m/s2
to = 0.74098462 s
6.23 or 6.24 hardly matters in my case.
You are given data to three significant figures. Standard is that you quote an answer to the same, so it would seem that you should give 62.4 as the answer.
But when the calculation involves taking differences this can go wrong. In the present case, replacing the 24.4 with 24.44 gives an acceleration of 6.26 m/s2. So the really correct way to answer would be something like 6.24±0.02.
 

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