Uniformly Accelerated Linear Motion

Straight Motion With Regular Change

Uniformly accelerated linear motion is motion along a straight path with constant acceleration. Constant acceleration means velocity changes by the same amount during each equal time interval.

If a car starts at $0\text{ m/s}$ and gains $2\text{ m/s}$ every second, its velocity becomes $2$ , $4$ , $6$ , and so on. The change is regular, not random.

In uniformly accelerated motion, velocity is not constant. Acceleration is constant.

v_t=v_0+at

Here, $v_0$ is initial velocity, $v_t$ is velocity after time $t$ , $a$ is acceleration, and $\Delta x$ is displacement.

Position Marks No Longer Stay Evenly Spaced

In uniform linear motion, the position marks each second are equally spaced. In uniformly accelerated motion, those gaps change. If the object speeds up, the gaps grow. If the object slows down, the gaps shrink.

Look at the "From Rest" motion. At the first instant the car is not moving yet, then the gaps between marks become longer. That is the sign that velocity is increasing. The graph rises in a straight line because the acceleration is constant.

Position Marks on a Rail

Marks on the rail show the train position every

1\text{ s}

while acceleration stays constant.

Initial velocity: $v_0=0\text{ m/s}$
Acceleration: $a=+4\text{ m/s}^2$
Final velocity: $v_t=+20\text{ m/s}$
Displacement: $\Delta x=50\text{ m}$

Displacement Comes from Graph Area

A velocity-time graph does more than show velocity. The area under the graph represents displacement. In uniformly accelerated motion, that area can be a triangle or a trapezoid.

Area Under the Velocity Graph

Choose the same motion, then read the area under the line as the object's displacement.

Situation	Area shape on the graph	Displacement formula
Starts from rest	Triangle	$\Delta x=\frac{1}{2}at^2$
Has initial velocity	Trapezoid	$\Delta x=\frac{1}{2}(v_0+v_t)t$
Uses acceleration directly	Rectangle plus triangle	$\Delta x=v_0t+\frac{1}{2}at^2$

These formulas are not separate tricks to memorize. They all come from the same idea: displacement is the area under a graph of $v$ against $t$ .

Car Starts from Rest

A toy car starts from rest, then has constant acceleration $2\text{ m/s}^2$ for $5\text{ s}$ .

Its final velocity is:

\begin{aligned} v_t &= v_0+at \\ &= 0+2(5) \\ &= 10\text{ m/s} \end{aligned}

Its displacement is:

\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2 \\ &= 0(5)+\frac{1}{2}(2)(5^2) \\ &= 25\text{ m} \end{aligned}

So after $5\text{ s}$ , the car is moving at $10\text{ m/s}$ and has moved $25\text{ m}$ from its starting point.

The two results answer different questions. Final velocity tells how fast the car is moving at the end, while displacement tells how far its position has changed during the whole interval.

Acceleration Sign Shows Change

In one-dimensional motion, positive and negative signs depend on the positive direction we choose. If rightward is positive, positive acceleration means the velocity change points rightward.

Negative acceleration does not automatically mean the object moves left. A car can still move right while having negative acceleration if it is braking. Its velocity points right, but its value gets smaller.