The acceleration is defined as the change in velocity with respect to time. If an object starts at rest with a velocity of zero m/sec and goes to a velocity of 1.5 m/sec in 1 second the acceleration would be 1.5 m/sec² (1.5 meters per second squared) or 1.5 m/sec per second. If the acceleration is positive the velocity is increasing. If the acceleration is negative the velocity is decreasing and we could say that the object is decelerating.
Friction always results in a negative acceleration. Friction slows objects down. The friction provides an outside force on the object that changes the object's velocity. If friction is the only outside force on the object an acceleration will occur (in this case a negative acceleration or deceleration). The greater the frictional force is, the greater the change in velocity will be and the greater the acceleration. In this case, the faster it will slow down.
When acceleration is present the final velocity will not equal the initial velocity. The final velocity will depend not only on the initial velocity, but on the acceleration as well. This is expressed in the following equation which is an extension of the equation from the last section when no acceleration was present (vf = vi).
vf = vi + a*t or vf = vi + at
Where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. The equation says that the final velocity is the initial velocity plus the acceleration times the time. The acceleration times the time accounts for the change in velocity due to acceleration. If the acceleration is zero the expression reduces to the former expression for no acceleration where the final velocity is equal to the initial velocity (the velocity is constant). It is a general strategy and always a good idea to make sure our equations make sense at the limits. In this case when the acceleration is zero. What happens at the limit when time is zero?
The expression for the distance traveled must also be modified from d = vt to incorporate the acceleration. You can use the last expression and the average velocity to derive the following expression:
d = vi*t + ½*a*t² or d = vit + ½at²
Where d is the distance, vi is the initial velocity, a is the acceleration, and t is the time. The equation says that the distance equals the initial velocity times the time plus one-half times the acceleration times the time squared. One-half times the acceleration times the time squared is the factor that accounts for acceleration. When the acceleration is zero the expression reduces back to distance equals the rate times the time. Also note that if no time elapses the distance traveled will be zero. This is clearly what should happen, but we must check our equations to be sure they really reflect the real world.
Application: How fast were they going before the crash?
As an example consider a policeman investigating an accident. If a car left skid marks that were 61 m long on a flat dry concrete road, how fast was the car moving before the brakes were applied? A car sliding on dry concrete has an acceleration of -6.86 m/sec² (this number is from experiment).
We are trying to determine the initial velocity. The final velocity is zero and the acceleration is -6.86 m/sec². First use vf = vi + at and solve for the initial velocity.
0 m/sec = vi + (-6.86 m/sec²)t or vi = (6.86 m/sec²)t
Unfortunately the time is not known. This expression for vi, however, can be used in d = vit + ½at² to determine the time.
61 m = vit + ½(-6.86 m/sec²)t2
61 m = (6.86 m/sec²)t² + ½(-6.86 m/sec²)t²
61 m = ½(6.86 m/sec²)t2
t = [(2)(61 m)/(6.86 m/sec²)]½ = 4.2 seconds
Note: the square root of a number is the same as the number raised to the ½ power. Now the time can be used in the previous equation to determine the initial velocity.
vi = (6.86 m/sec²)t = (6.86 m/sec²)(4.2 sec) = 28.9 m/sec (about 65 mi/hr)
Another Option. That was a little complicated. Policemen actually have a third equation of motion that they can use which gets the result more directly as shown below.
The third equation is:
vf² = vi² + 2ad
In words it says that the final velocity squared is equal to the initial velocity squared plus two times the acceleration times the distance traveled. Solving for the initial velocity gives:
vi² = vf² - 2ad
Taking the square root of both sides results in the following equation that we can use to solve for the initial speed of the car.
vi = [vf² - (2)(a)(d)]½ = [0 - (2)(-6.86m/sec²)(61 m)]½ = 28.9 m/sec
Extension. Try this simpler problem. A wooden box was on a conveyer belt that moved the box at a constant speed. The end of the conveyer belt connected to a level stone floor where the box would slide and come to rest. The workers have 3 seconds from the time the box hits the floor to the time when it stops. Wood on stone has an acceleration of -3.92 m/sec². What was the velocity of the wooden box when it first hit the stone floor? (Answer: 11.76 m/sec)