Driver reaction time, with negative acceleration

[Spydermonkey]

Homework Statement

The "reaction time" of the average automobile driver is about 0.700 . (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of 12.0 , compute the total distance covered in coming to a stop after a signal is observed (a) from an initial velocity of 15.0 (in a school zone) and (b) from an initial velocity of 55.0 .
x=?
xo=?
vx=0, point at which you are stopped
vxo=15 mi/h initial speed
ax=-12 ft/s^2
t=?

Homework Equations

v=d/t
vx^2=vxo^2+2ax(x-xo)

The Attempt at a Solution

I am pretty much lost, I tried converting everything to meters as that is what the answers call for, and got the equation

vx^2-vxo/2ax= (0)^2 -(6.7056)^2/ 2(-3.6576)= 6.1468 meters

 

[cristo]

Take x0=0, split the motion into two parts; before the driver brakes, and when the driver brakes and thus decelerates. You can calculate the distance for the first part with your first equation, and the second part with your second equation.

 

[Spydermonkey]

Take x0=0, split the motion into two parts; before the driver brakes, and when the driver brakes and thus decelerates. You can calculate the distance for the first part with your first equation, and the second part with your second equation.

I think the first part is not meant to be figured out, as it is the time before the signal was noticed, and I am supposed to measure only the second portion? Or am I just completely confused about what it is asking?

 

[cristo]

I meant that you have two parts contributing to the distance.
1. The signal has been noticed, and the driver is responding. Here the car is traveling at constant velocity, and so the distance can be found through x=vt.
2. The driver has applied the brakes. Here, the vehicle is declerating constantly and you can use the formula v²=v₀²+2ax to find the distance.

 

[Spydermonkey]

Thanks, I will try plugging my values into those equations.

 

[Spydermonkey]

crud, I don't know t though.

 

[cristo]

For the first equation, t will be the reaction time of the driver. The second part does not need time.

 

[Spydermonkey]

So basically x=10.5 mi/h/s or 4.69 meters/s^2.

 

[cristo]

Your units are not correct. The units for distance will be either miles or metres. Convert everything into metres, seconds, and metres per second before you plug into equations.

 

[Spydermonkey]

When I converted I got 4.69 meters per second squared.

 

[cristo]

Your numerical answer is correct, but your units are not. Note that $$x=vt\Rightarrow [m]=[m/s]$$ Where [ ] denotes units. Note that the units of time cancel out on the right hand side, and so the units for distance are metres.

 

[Spydermonkey]

I will have at it for awhile, if I have any other questions I will pop in again, thanks.

 

[leahtiff]

a motorist is traveling at 18 m/s when he sees a deer in the road 49 m ahead. if the maximum negative acceleration of the vehicle is -7m/s^2, what is the maximum reaction time (t) of the motorist that will allow him to avoid hitting the deer?

If his reaction time is 1.62541 s, how fast will he be traveling when he reaches the deer?

 

[leahtiff]

a motorist is traveling at 18 m/s when he sees a deer in the road 49 m ahead. if the maximum negative acceleration of the vehicle is -7m/s^2, what is the maximum reaction time (t) of the motorist that will allow him to avoid hitting the deer?

If his reaction time is 1.62541 s, how fast will he be traveling when he reaches the deer?the original formula i s used is: vf^2 = vi^2+2ad and (a-a)/d=t