Equations of Motion: Solving "Reaction Time" Homework

[jackleyt]

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 .

Homework Equations

Equations of Motion

The Attempt at a Solution

v (not) =6.71 m/s
v (final) = 0 m/s
t=.700 s
a= -3.6576 m/s^2
x (final) = variable
x (not) = 0

x(final) = x(not) + v(not)t+(1/2at^2)
x(final) = 0 + 6.71(.7) +(1/2*-3.6576)(.7)^2
x(final) = 1.1126 meters

What did I do wrong?

 

[LowlyPion]

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 .

Homework Equations

Equations of Motion

The Attempt at a Solution

v (not) =6.71 m/s
v (final) = 0 m/s
t=.700 s
a= -3.6576 m/s^2
x (final) = variable
x (not) = 0

x(final) = x(not) + v(not)t+(1/2at^2)
x(final) = 0 + 6.71(.7) +(1/2*-3.6576)(.7)^2
x(final) = 1.1126 meters

What did I do wrong?

Welcome to PF.

The .7 seconds is the reaction time. All that means is that there is a delay from when the driver observes the signal and when the deceleration begins. Hence to find the Total distance for each case just use the x = .7*V as the distance traveled before application of the brakes, and then use the V² = 2*a*x to determine the remainder of the stopping distance.

 

[baba89]

Your approach is correct, but there is a minor error in your calculation. When solving for the total distance covered, you should use the absolute value of the acceleration (since it is a scalar quantity) instead of the negative value. So, the correct equation should be:

x(final) = x(not) + v(not)t + (1/2|a|t^2)

Using this equation, the total distance covered for both scenarios would be:

(a) x(final) = 0 + 6.71(0.7) + (1/2*12*0.7^2) = 8.34 meters
(b) x(final) = 0 + 6.71(0.7) + (1/2*12*0.7^2) = 24.24 meters

So, the total distance covered in coming to a stop after a signal is observed from an initial velocity of 15.0 m/s is 8.34 meters, and from an initial velocity of 55.0 m/s is 24.24 meters.