Traveling at a speed of 15.6 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.730. What is the speed of the automobile after 1.25 s have elapsed? Ignore the effects of air resistance.

Vf=Vi+at where a= forcefriction/mass= .730*mg/m= .730g

I don't have mass so I can't use this formula :(

good grief.

.730mg divided by m is .730g

No matter how I use the formula you provided I do not get the correct answer of: 6.66 m/s.

You may want to drop physics.

Put this into your google search window
15.6-.730*9.8*1.25=
which is
Vf-mu*g*1.25

Thanks for your help bobpursley. However, your advise was unwarranted. Just because someone needs additional help understanding a concept doesn't mean they should "drop" out of a class. Jiskha is a learning environment for those seeking help to better understand ideas and concepts please remember that.

To find the speed of the automobile after 1.25 seconds, we can use the equation of motion:

vf = vi + at

Where:
vf is the final velocity,
vi is the initial velocity,
a is the acceleration, and
t is the time.

Given information:
Initial velocity (vi) = 15.6 m/s
Time (t) = 1.25 s

Now, we need to find the acceleration (a) in order to plug the values into the equation.

The acceleration can be determined using the equation:

a = μ * g

Where:
μ is the coefficient of kinetic friction,
g is the acceleration due to gravity (approximately 9.8 m/s²).

Given information:
Coefficient of kinetic friction (μ) = 0.730

Substituting the given values, we get:

a = 0.730 * 9.8

Calculating the value of 'a', we get:

a ≈ 7.154 m/s²

Now, we can substitute the values of 'vi', 'a', and 't' into the equation of motion:

vf = 15.6 + (7.154 * 1.25)

Calculating the value of 'vf', we get:

vf ≈ 24.57 m/s

Therefore, the speed of the automobile after 1.25 seconds is approximately 24.57 m/s.