Rectilinear Acceleration


Equations

a  =  
 
lim
Δt→0
Δv
Δt
 = 
dv
dt
 = 
d2x
dt2
  =  v
dv
dx
Instantaneous acceleration
Average acceleration =
Δv
Δt
Average acceleration


Acceleration

Consider the velocity u of the particle at time t and also its velocity u + Δu at a later time t + Δt.



The average acceleration of the particle over the time interval Δt is defined as the quotient of Δu and Δt:

Average acceleration =
Δu
Δt

The equation basically says how fast the particle went from one velocity to another velocity. If something went from 5 mph to 10 mph in 5 seconds (approximately 0.00139 hours), then the velcity increased at a rate of 3600 miles/hour2 (or 1 mile/min2). Because the length units are squared, it can create this illusion of 'too large' or 'too small' answers.

The instantaneous acceleration a of the particle at the instant t is obtained from the average acceleration by choosing smaller and smaller values for Δt and Δu:

Instantaneous acceleration = a =
 
lim
Δt→0
Δu
Δt

The limit of the quotient, which is by definition the derivative of u with respect to t, measures the rate of change of the velocity:

(Eq2)    
a =
du
dt

Or, substituting for u from Eq1 of the lesson Rectilinear Velocity,

(Eq3)    
a =
d2x
dt2

The acceleration a is represented by an algebraic number which can be positive or negative. A positive value of a indicates that the velocity (that is, the algebraic number u) increases. This may mean that the particle is moving faster in the positive direction



or that it is moving more slowly in the negative direction:



in both cases, Δu is positive. If a ball is ejected vertically into the air, at a velocity u, and the acceleration is equal to -9.8 m/s2 (assuming the acceleration is equal to earth's gravity at sea level and the positive sense is upward), then while the ball is going up, it is moving more slowly in the positive direction (decelerating, the velocity is decreasing) and when the ball starts to come back down, it starts to move faster in the negative direction (still decelerating, the velocity is still decreasing (velocity is growing larger in the negative direction)). Either way, when the ball is going up, or down, the gravity is the same and the acceleration is equal to -9.8 m/s2. A negative value of a indicates that the velocity decreases; either the particle is moving more slowly in the positive direction



or it is moving faster in the negative direction.



The term deceleration is sometimes used to refer to a when the speed of the particle (that is, the magnitude of u) decreases; the particle is then moving more slowly. For example, the particle in the following figure is decelerated in parts b and c; it is truly accelerated (that is, moves faster) in parts a and d.

Another expression for the acceleration can be obtained by eliminating the differential dt in Eq1 and Eq2. Solving Eq1 for dt, dt = dx/u is obtained; substituting into Eq2:

a = u
du
dx