How far are you from the intersection when you begin to apply the brakes?

[osker246]

Homework Statement

You are diving to the grocery store at 20 m/s. You are 110 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.70 s and that your car brakes with constant acceleration.

A) How far are you from the intersection when you begin to apply the brakes?
B) What acceleration will bring you to rest right at the intersection?
C)How long does it take you to stop?

Homework Equations

Xf = Xi + ((Vx)i)(Delta T) + .5(Ax)(Delta T)^2

The Attempt at a Solution

This may be a pretty basic problem for you folks here. But I honestly don't have the slightest clue where to even start. I'd imagine id have to find the acceleration of the vehicle in order to find part A. But I don't see how to find it. Ax= Delta Vx/ Delta T correct? So with the information given would it be Ax= 20m/s / 0.70 s ?

I have the answers to the problem from my textbook. I Just don't understand the steps in solving for the answer. If anybody can give me a bump in the right direction I would appreciate it. Thanks.

Oh yeah, Answers are

A) 10 m
B) -2 m/s^2
C) 11 sec

 

[tiny-tim]

Welcome to PF!

You are diving to the grocery store at 20 m/s. You are 110 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.70 s and that your car brakes with constant acceleration.

A) How far are you from the intersection when you begin to apply the brakes?
…
I'd imagine id have to find the acceleration of the vehicle in order to find part A.
…
I Just don't understand the steps in solving for the answer. If anybody can give me a bump in the right direction I would appreciate it. Thanks.

Hi osker246! Welcome to PF!

(I do nudges rather than bumps … )

Part A is just distance = speed x time … you don't need acceleration at all!

In fact, look at your equation:

Ax= 20m/s / 0.70 s ?

(m/s) / s would be m/s², wouldn't it?

Always check the dimensions of a formula, if you're not sure.

 

[osker246]

Hi osker246! Welcome to PF!

(I do nudges rather than bumps … )

Part A is just distance = speed x time … you don't need acceleration at all!

In fact, look at your equation:


(m/s) / s would be m/s², wouldn't it?

Always check the dimensions of a formula, if you're not sure.

Thank you for the welcome!

I don't know if D=RT is a proper application to this problem though. The only time given in this problem is .70 sec which refers reaction time not travel time. If I'm incorrect please do inform me. I'm just a bit lost where to go with this problem.

 

[osker246]

ok I am just messing around with numbers right now.

So 110 m / 20m/s = 5.5 sec

So if the vehicle continues at a rate of 20m/s it will take 5.5 sec to travel the 110 m correct?

So finding the acceleration should be as easy as 20 m/s / 5.5 s = 3.63 m/s^2. Right?

So would I go about using this equation?

Xf = Xi + ((Vx)i)(Delta T) + .5(Ax)(Delta T)^2

xf=110 m
(Vx)i=20 m/s
Delta T= .70 sec?
Ax= 3.63

Then solve for its initial position?

 

[osker246]

yeah...I have no idea what the hell I'm doing...

 

[cryptoguy]

Thank you for the welcome!

I don't know if D=RT is a proper application to this problem though. The only time given in this problem is .70 sec which refers reaction time not travel time. If I'm incorrect please do inform me. I'm just a bit lost where to go with this problem.

Yea, you can use d = rt for this because that means that for .7 secs you're still traveling at a constant rate of 20 m/s.

For B, use the distance you found in A; remember, you also know the initial velocity, the final velocity (what equation can you use?)

For C, you know acceleration, initial velocity and final velocity (what equation will work here?)

 

[osker246]

Yea, you can use d = rt for this because that means that for .7 secs you're still traveling at a constant rate of 20 m/s.

So...

D=(.70)(20)=14 m

for that .70 of a second the vehicle traveled 14 m. So 110m - 14m = 96m

So does that mean I'm 96m away from the intersection before starting to slow down?

 

[elessariitkgp]

Let the distance from the intersection, when the driver sees the red light be L, let his rection time be t_{0}. Let the initial speed of the car be v_{0}. If the driver decides to put on the brakes the moment he sees the red light, the action still takes time t_{0}, and in that time the car has traveled a distance v_{0}t_{0}. Hence the car travels with deceleration for the distance L-v_{0}t_{0}. This is the answer to (a)
If a be the magnitude of the deceleration, then
0=v^{2}_{0}-2a(L-v_{0}t_{0})
which gives a, the answer to (b)
The time can be found from
L-v_{0}t_{0}=v_{0}T-\frac{1}{2}aT^{2}
which gives the time of deceleration. The time of stopping will be T+t_{0}, which gives the answer to (c)

 

[cryptoguy]

So...

D=(.70)(20)=14 m

for that .70 of a second the vehicle traveled 14 m. So 110m - 14m = 96m

So does that mean I'm 96m away from the intersection before starting to slow down?

Correct

 

[tiny-tim]

reaction time

Hi osker246!

(I've been away all day, but I see cryptoguy has helped you to the right result for A).)

I don't know if D=RT is a proper application to this problem though. The only time given in this problem is .70 sec which refers reaction time not travel time. If I'm incorrect please do inform me. I'm just a bit lost where to go with this problem.

Time is time. The .70 sec is both reaction time and travel time.

Reaction time (you don't drive do you?) means the time it takes someone (usually between .60 and 1.5 sec, I think) to react after they see something … as when you see someone lying in the road, and you "immediately" slam on the brakes, but it takes .70 sec for the message to get from you eyes to your brain, and then back to your foot!

In other words: it takes you .70 sec just to "react", before you actually do anything!

(If you're not sure what a word means, just ask! )

How are you doing on B) and C)?

 

[osker246]

I got the problem solved. Turned out the teacher handed out the wrong answer for part A. This is why I had so much trouble trying to solve this problem. My answers were not matching the one he originally handed out. I just want to say thank you for the help everybody! Its greatly appreciated.