• Motion  ▼

• Linear motion  ▼

# Uniformly Accelerated Motion in a Striaght Line

When the acceleration is constant and its direction is parallel to the initial velocity, the object moves along a straight line with uniform acceleration. It is a special example of accelerated motion.

## Uniformly accelerated linear motion: velocity, acceleration and time

$a=\frac{\mathrm{\Delta v}}{\mathrm{\Delta t}}=\frac{\mathrm{v - v₀}}{\mathrm{t - t₀}}⇒$

$\mathrm{\Delta v}=\mathrm{a\Delta t}⇒$

$v=\mathrm{v₀ + a\Delta t}⇒$

$v=\mathrm{v₀ + at}$

You can make the formula more concise by replacing Δt with t. Since the acceleration is constant, the velocity-time graph is a sloping line, or it is a linear function. A straight line sloping downward indicates a negative direction of acceleration and it is opposite to the positive direction you prescribed. A straight line sloping upward indicates a positive direction of acceleration.

When the acceleration is in the same direction as the initial velocity, the object accelerates. When the acceleration is in the opposite direction of the initial velocity, the object decelerates and then accelerates in the opposite direction.

## Uniformly accelerated rectilinear motion: displacement, initial velocity, acceleration and time Since the acceleration is constant, the displacement-time graph is a quadratic curve. Its tangent line represents the velocity.

The concave curve ⇔ a ＜ 0, the acceleration direction is opposite to the x-axis. The convex curve ⇔ a > 0, the acceleration direction is the same as the x-axis. The displacement is the area enclosed by the curve and t axis in the v-t graph. This is equivalent to solving for the area of a trapezoid.

$\mathrm{x - x₀}=\frac{\mathrm{\left(v + v₀\right)t}}{2}\mathrm{·····························①}$

$v=\mathrm{v₀ + at}$Replace v in equation ① with this.

$\mathrm{x - x₀}=\mathrm{v₀t +}\frac{\mathrm{at²}}{2}\mathrm{·····························②}$

Both sides of formula ① divided by t can get average velocity in the time interval of t.

$\mathrm{Average velocity}=\frac{\mathrm{x - x₀}}{t}=\frac{\mathrm{v + v₀}}{2}$

The average velocity is also equal to the velocity at the intermediate moment.

$\frac{\mathrm{v + v₀}}{2}=\frac{\mathrm{v₀ + v₀ + at}}{2}=\mathrm{v₀ +}\frac{\mathrm{at}}{2}={V}_{t/2}^{}$

## Uniformly accelerated linear motion: displacement, velocity and acceleration

$\mathrm{x - x₀}=\frac{\mathrm{\left(v + v₀\right)t}}{2}$,$t=\frac{\mathrm{\left(v - v₀\right)}}{a}$

Eliminating t results in a formula for displacement and velocity.

v² - v₀² = 2a（x - x₀）

## Velocity of one half displacement

An object moves from A to B with uniform acceleration, find the velocity at the midpoint of AB. The velocity of the object at A is v₀, the velocity at B is v₁, the velocity at the midpoint of AB is v , the displacement x - x₀ from A to B, and the time interval between A and B is t .

$\mathrm{v² - v₀² = 2a（x - x₀） ÷ 2 = a\left(x - x₀\right)= a t}\stackrel{—}{v}$

$\mathrm{at = v₁ - v₀,}\stackrel{—}{v}=\frac{\mathrm{v₁ + v₀}}{2}\mathrm{Plug these two formulas into the first one.}$

$\mathrm{v² - v₀² = \left(v₁ + v₀\right)\left(v₁ - v₀\right) ÷ 2}⇒\mathrm{v² = \left(v₀² + v₁²\right)/2}$

## The acceleration of most motions is not constant, so what is the significance of studying linear motion with constant acceleration?

It is the simplest non uniform motion, so its details are the easiest to explore clearly.

There are also some common movements that belong to uniformly accelerated motion, such as throwing objects upward or downward, and free fall.

In many cases, we do not need to study every detail of motion accurately like scientists. We only want to quickly and semi-quantitatively draw conclusions, so it is meaningful to approximate real motion as uniformly accelerated motion. For example, the start of a car and a bullet entering a wooden block can be regarded as motion with constant acceleration.

## How can the problem of uniformly accelerated linear motion be solved?

Whether the velocity, displacement and acceleration in these formulas are positive or negative will be determined by the coordinate system you build. If you need to plug values into the equations, the first step should be to set up a coordinate system to determine the sign of the values. Generally, these types of problems are relatively simple and can be directly calculated by plugging the numbers into the formula. The more difficult problems require writing the object's displacement function with respect to time.

## When will A and B meet?

Object A with constant velocity is chasing object B which is moving with constant acceleration. The distance between them is s. The velocity of object A is vₐ , while the initial velocity of object B is v₀ and acceleration of B is in the same direction of the initial velocity.

Since all velocities and acceleration are the in the same directions, we can assume that the positive direction of the coordinate system is the same as the direction of velocity.

Xₐ = vₐt

Xᵦ = s + v₀t + at²/2

F(t) = Xᵦ - Xₐ= s+ (v₀- vₐ)t + at²/2

When F(t) = 0, A meets B. This is equivalent to solving a quadratic equation. The equation has a solution when Δ = (v₀- vₐ)² - 2as ≥ 0.