
The equation defining acceleration can be rewritten as a model that describes how the velocity changes over time. The result is equation (4.2), which gives the instantaneous velocity v at time t under the assumption that the acceleration a is constant.

In equation (4.2), we are not using the notation Δt = t_{f} − t_{i}. Instead, time t is used as a continuously variable quantity. When we make this change for time t, the way we write the velocity also changes. In equation (4.2) we are creating a model for the velocity that can be used for any time t. So we use the notation that v_{0} is the velocity when time is t = 0, but we use v to represent the velocity at any time. Many people will use the notation v(t) to represent explicitly how velocity is a function of time.
An alternate way to express equation (4.2) for two particular points in time is $${v}_{f}={v}_{i}+a\text{\Delta}t={v}_{i}+a({t}_{f}{t}_{i})$$
 This equation, however, represents the velocity at two particular times and is not a model for the velocity.

(4.2)  $$v={v}_{0}+at$$
 v  =  velocity (m/s)  v_{0}  =  initial velocity (m/s)  a  =  acceleration (m/s^{2})  t  =  time (s) 
 Velocity for constant acceleration 

In this equation we set Δt = t by assuming the motion starts at t = 0. The initial velocity v_{0} is the velocity when t = 0.

The velocity model in equation (4.2) corresponds to the equation of a straight line when plotting velocity versus time. In recognizing this comparison between the mathematics of a straight line and the velocity in accelerated motion, don’t be tricked because the righthand side of equation (4.2) has the position of the terms at and v_{0} switched! The slope of the line—the coefficient in front of time t—is the acceleration a. The intercept on the vertical axis—the constant value added to the equation—is the initial velocity v_{0}.

A cart traveling at 1 m/s reaches a hill and accelerates down the hill at 0.5 m/s^{2}. What is the velocity of the cart 3 s after it starts accelerating?

Asked: 
instantaneous velocity v 
Given: 
initial velocity of v_{0} = 1 m/s, acceleration a = 0.5 m/s^{2}, and time t = 3.0 s 
Relationships: 
v = v_{0} + at 
Solution: 
v = 1 m/s + (0.5 m/s^{2})(3.0 s) 

= 
2.5 m/s 

In the above example the velocity and acceleration are in the same direction—both are positive. The speed increases from 1 to 2.5 m/s. Acceleration can also decrease an object’s speed when the sign of the velocity is different from the sign of the acceleration. For example, an acceleration of −1 m/s^{2} adds −1 m/s to the velocity each second. This would decrease a positive velocity. It would also make a negative velocity more negative—that is, faster in the negative direction.

A cart traveling at 2 m/s along a level surface reaches an upward sloping hill and accelerates at −0.5 m/s^{2}. What is the velocity of the cart 3 s after it starts climbing the hill?

Asked: 
instantaneous velocity v 
Given: 
initial velocity of v_{0} = 2 m/s, acceleration of a = −0.5 m/s^{2}, and time of t = 3 s 
Relationships: 
v = v_{0} + at 
Solution: 
v = 2 m/s + (−0.5 m/s^{2})(3.0 s) 

= 
0.5 m/s 

The model of motion given by equation (4.2) applies to the instantaneous velocity at time t. Translated to an English sentence the equation tells us that the velocity v at time t is the initial velocity v_{0} plus the change in velocity due to acceleration a applied every second for t seconds. The assumption of constant acceleration means that the change in velocity each second is the same.

Can you describe a situation in which an object’s acceleration is negative but its speed is increasing?

If the velocity is negative, then a negative acceleration makes the velocity more negative every second. This increases the speed since speed is the absolute value of velocity. For example, suppose the velocity starts at −10 m/s and the acceleration is −1 m/s^{2}. After one second the velocity is −11 m/s. After two seconds the velocity is −12 m/s. The speed increases from 10 to 12 m/s.
