Estimating Car Speed from Skid Marks

[nordqvist11]

Homework Statement

In coming to a stop a car leaves skid marks 92m long on the highway. Assuming a decceleration of 7.00 m/s^2 , estimate the SPEED of the car just before braking.

I have a question though, is my approach ok? You can see my confusion in various parts of my procedures below, I guess it's the fact that it says find speed not velocity.

Homework Equations

Kinematic Equation for constant acceleration.

The Attempt at a Solution

 

[iRaid]

Speed=velocity lol

 

[nordqvist11]

I got -35m/s as my velocity though

 

[lachy89]

Speed is the magnitude of velocity and has no direction.

EDIT: Assuming you are using the formula: v^2=u^2+2as you may need to check what you are using as v and u and be careful of your signs.

0 = u^2 +2as
u^2=-2as

Does that make sense?

 

[rwisz]

I understand your confusion about the difference between speed and velocity. In this scenario, we are looking for the speed of the car, which is the magnitude of its velocity. The kinematic equation for constant acceleration can be used to solve for the initial velocity of the car before braking.

First, we need to rearrange the equation to solve for initial velocity (v0):

v0 = vf - at

Where:
v0 = initial velocity (unknown)
vf = final velocity (0 m/s since the car comes to a stop)
a = acceleration (-7.00 m/s^2)
t = time (unknown)

Next, we need to find the time it took for the car to come to a stop. We can use the equation d = v0t + 1/2at^2, where d is the distance (92m) and v0 is the initial velocity we are trying to solve for. Rearranging the equation, we get:

t = (-v0 ± √(v0^2 + 4ad)) / 2a

Plugging in the values, we get:

t = (-v0 ± √(v0^2 + 4(-7.00)(92))) / 2(-7.00)

Simplifying, we get two possible values for t:

t1 = (-v0 + 30.4) / -14.0
t2 = (-v0 - 30.4) / -14.0

Since we are looking for the time it took for the car to come to a stop, we can ignore the negative value for t2. Now, we can plug in t1 into our first equation to solve for v0:

v0 = 0 - (-7.00)(t1)
v0 = 7.00t1

Substituting t1, we get:

v0 = 7.00((-v0 + 30.4) / -14.0)

Solving for v0, we get:

v0 = 30.4 / 7.00 = 4.34 m/s

Therefore, the estimated speed of the car just before braking is 4.34 m/s. It is important to note that this is just an estimate and may not be completely accurate due to factors such as friction and other external forces. Further analysis and measurements may be needed to get a more precise