Study Notes on Motion with Constant Acceleration
Motion with Constant Acceleration
Focus Question
How are position, velocity, acceleration, and time related?
Position with Constant Acceleration
If an object experiences constant acceleration, its velocity changes at a constant rate.
Position changes with time:
The positions of a car experiencing constant acceleration are graphed.
The graph shows that the car's motion is not uniform.
Displacements for equal time intervals get larger as time progresses.
The slope of the line in the graph becomes steeper over time.
For constant acceleration, the position-time graph forms a parabola.
Position vs. Time
Graphical representation of position shows how it varies with time.
Example data points:
At $t=1.00$ s, position is $0.0$ m.
At $t=5.00$ s, position is $60.0$ m.
The slope between points gives velocity:
Between $t=1.00$ s and $t=3.00$ s, slope (velocity) is calculated as:
m = \frac{20.0 \text{ m} - 0.00 \text{ m}}{3.00 \text{ s} - 1.00 \text{ s}} = 10.0 \text{ m/s}Between $t=3.00$ s and $t=5.00$ s, another slope gives:
m = \frac{60.0 \text{ m} - 20.0 \text{ m}}{5.00 \text{ s} - 3.00 \text{ s}} = 20.0 \text{ m/s}
Slope and Area Under the Graph
The slope of a position-time graph indicates speed/velocity.
The area under the graph drives calculations of displacement.
Example velocity-time graph (derived from the position-time graph) displays constant acceleration as a straight line.
The rise divided by the run gives acceleration:
m = \frac{\Delta v}{\Delta t} = 5.00 \text{ m/s}²
Velocity with Average Acceleration
Formula for average acceleration manipulation:
a = \frac{\Delta v}{\Delta t}
Rearranged to find final velocity:
Final Velocity equation:
vf = vi + a \Delta tDescribes how the velocity of an object changes under constant acceleration.
Real-World Applications
Drag Racing Example:
Fastest recorded time for a dragster over a 402-m course: $3.771$ s.
Highest final speed: $145.3$ m/s (324.98 mph).
Practice Problems:
A golf ball rolls toward a hole:
Initial speed: $2.0$ m/s; slows at $0.50$ m/s².
Find velocity after $2.0$ s and $6.0$ s.
Describe the motion of the ball.
Finding Displacement from a Velocity-Time Graph
Calculate displacement from the area under a v-t graph.
Example: Given a constant velocity of $75$ m/s:
Displacement for $\Delta t = 1.0$ s:
\Delta x = v \Delta t = 75 \text{ m/s} \cdot 1.0 \text{ s} = +75 \text{ m}For $\Delta t = 2.0$ s:
\Delta x = v \Delta t = 75 \text{ m/s} \cdot 2.0 \text{ s} = +150 \text{ m}
Motion with Initial Non-Zero Velocity
For objects with initial velocity, the displacement involves calculating both rectangular and triangular areas under the v-t graph:
Total area:
A = A{rectangle} + A{triangle}If initial or final position known, position equation evolves to:
x - xi = vi t + \frac{1}{2} a t²Alternative velocity equation without time dependency:
v² = vi² + 2a(x - xi)
Example Problem: Displacement of an Accelerated Automobile
An automobile starts from rest ($v_i = 0$) and accelerates at $3.5$ m/s² to finally reach $25$ m/s:
Using the velocity-displacement relation: v² = vi² + 2a(x - xi)
Plugging in the parameters:
25² = 0 + 2 \cdot (3.5)(x)Solve for $x$, which gives a reasonable distance based on vehicle motion.
Practice Problems
A skateboarder slows down on an incline.
A race car goes from $44$ m/s to $22$ m/s in $11$ s, how far does it move?
A car accelerates from $15$ m/s to $25$ m/s over a distance of $125$ m, find the duration of the acceleration.
Challenge involving displacement, velocity, and acceleration relationships.
Additional Practice Problems
Set of realistic scenarios integrating understanding of velocity, time, acceleration, and displacement calculations based on previous lessons and principles.