# One-Dimensional Kinematics with Constant Acceleration| University Physics

Learning Goal:

To understand the meaning of the variables that appear in the equations for one-dimensional kinematics with constant acceleration.

Motion with a constant, nonzero acceleration is not uncommon in the world around us. Falling (or thrown) objects and cars starting and stopping approximate this type of motion. It is also the type of motion most frequently involved in introductory kinematics problems.

The kinematic equations for such motion can be written as

$x\left(t\right)=x_i+v_it+\frac{1}{2}at^2$

$v\left(t\right)=v_i+at,$

where the symbols are defined as follows:

• $x\left(t\right)$ is the position of the particle;
• $x_i$ is the initial position of the particle;
• $v\left(t\right)$ is the velocity of the particle;
• $v_i$ is the initial velocity of the particle;
• $a$ is the acceleration of the particle.

In answering the following questions, assume that the acceleration is constant and nonzero: a≠0.

### PART B. The quantity represented by $x_i$ is a function of time (i.e., is not constant).

Recall that $x_i$ represents an initial value, not a variable. It refers to the position of an object at some initial moment.

### PART D. The quantity represented by v is a function of time (i.e., is not constant).

The velocity v always varies with time when the linear acceleration is nonzero.

### PART E. Which of the given equations is not an explicit function of t and is therefore useful when you don’t know or don’t need the time?

ANSWER: $v^2=\left(v_i\right)^2+2a\left(x-x_i\right)$

### PART F. A particle moves with constant acceleration a. The expression vi+at represents the particle’s velocity at what instant in time?

ANSWER: when the time t has passed since the particle’s velocity was $v_i$

More generally, the equations of motion can be written as

$x\left(t\right)=x_i+v_i\Delta t+\frac{1}{2}a\left(\Delta t\right)^2$

and

$v\left(t\right)=v_i+a\Delta t$

Here Δt is the time that has elapsed since the beginning of the particle’s motion, that is, $\Delta t=t-t_i$, where $t$ is the current time and $t_i$ is the time at which we start measuring the particle’s motion. The terms $x_i$ and $v_i$ are, respectively, the position and velocity at $t=t_i$. As you can now see, the equations given at the beginning of this problem correspond to the case $t_i=0$, which is a convenient choice if there is only one particle of interest.

### What is the equation describing the position of particle B?

ANSWER: $x_B\left(t\right)=x_i+0.5v_i\left(t-t_1\right)+a\left(t-t_1\right)^2$

The general equation for the distance traveled by particle B is $x_B\left(t\right)=x_{iB}+v_{iB}\Delta t+\frac{1}{2}a_B\left(\Delta t\right)^2$ or $x_B\left(t\right)=x_{iB}+v_{iB}\left(t-t_1\right)+\frac{1}{2}a_B\left(t-t_1\right)^2$, since $\Delta t=t-t_1$ is a good choice for B. From the information given, deduce the correct values of the constants that go into the equation for $x_B\left(t\right)$ given here, in terms of A’s constants of motion.

$x_B\left(t\right)=x_i+\frac{1}{2}v_i\left(t-t_1\right)+\frac{1}{2}\left(2a\right)\left(t-t_1\right)^2$

$x_B\left(t\right)=x_i+0.5v_i\left(t-t_1\right)+a\left(t-t_1\right)^2$

### PART H. At what time does the velocity of particle B equal that of particle A?

ANSWER: $t=2t_1+\frac{v_i}{2a}$

Particle A’s velocity as a function of time is $v_A\left(t\right)=v_i+at$, and particle B’s velocity as a function of time is $v_B\left(t\right)=0.5v_i+2a\left(t-t_1\right)$.

Once you have expressions for the velocities of A and B as functions of time, set them equal and find the time t at which this happens.

$v_i+at=\frac{1}{2}v_i+2a\left(t-t_1\right)$

$v_i+at=\frac{1}{2}v_i+2at-2at_1$

$2at-at=2at_1+\frac{1}{2}v_i$

$at=2at_1+\frac{1}{2}v_i$

$t=2t_1+\frac{v_i}{2a}$