Unit 1 Kinematics Flashcards
A car accelerates uniformly from rest. Which of the following statements is true?
A. Its displacement is directly proportional to time.
B. Its displacement is proportional to the square of the time elapsed.
C. Its velocity is constant throughout.
D. Its acceleration decreases with time.
Recognize that under constant acceleration, the displacement equation
𝑥=𝑥0+𝑣0𝑡+1/2𝑎𝑡^2
Implies a quadratic relationship between displacement and time.
Correct Answer: B
(This example shows how conceptual understanding of the time-squared dependence is tested.)
Question:
A ball is thrown upward with an initial velocity 𝑣0. Derive an expression for the maximum height H reached by the ball.
1.Start with the equation
𝑣^2=𝑣0^2+2𝑎(𝑥−𝑥0)
- v =0 and a=-g
3. Rearranging
0=v0^2−2gH⟹H= V0^2/2g
At maximum height, the final velocity
Clearly explain each step and state assumptions (e.g., neglecting air resistance).
(This FRQ emphasizes the derivation process and clear presentation of reasoning.)
What does the phrase “maximum height” indicate in a projectile motion problem?
It indicates the point where the vertical component of velocity becomes zero (v = 0). This is used with equations such as
𝑣^2=𝑣0^2+2aΔy to solve for height.
Define “uniform acceleration” and its significance in kinematics.
Uniform acceleration” means the acceleration is constant over time. This allows the use of standard kinematic equations like
𝑥=𝑥0+𝑣0t+1/2𝑎𝑡^2 without needing to integrate a variable acceleration.
How do you differentiate between “displacement” and “distance”?
Displacement is a vector quantity (it includes direction) that represents the change in position, while distance is a scalar that indicates the total length of the path traveled.
(MCQ Scenario) A ball is thrown upward. Which statement is true at the ball’s maximum height?
Its vertical velocity is zero. Recognize keywords like “maximum height” to set
𝑣=0
(FRQ Scenario) Derive an expression for the maximum height reached by an object thrown upward with initial speed v0
.
A: Use
𝑣^2=𝑣0^2+2(−𝑔) 𝐻 with v=0
Solve for
H: 𝐻=𝑣0^2/2𝑔
(“Derive” and “maximum height” cue the use of boundary condition v=0)
If you go ski area, you likely find that the beginners slope has the smallest angle. Use the concept of acceleration on a ramp to explain why this is so.
ax=-gsin(angle)
Maximum Height / At the Apex:– Keywords: “maximum height,” “apex,” “peak,” “highest point.” means what?
Implication: The vertical (final) velocity becomes zero at the peak; use
𝑣𝑓2=𝑣02+2𝑎Δ𝑦 with v0=0
solve for height.
“uniform,” “constant acceleration.”
You may apply the kinematic equations without worrying about variable acceleration.
“A ball is thrown vertically upward. At its maximum height, its velocity is …”
Keywords: “maximum height” signal
V=0
– Answer Strategy: Recognize that kinetic energy is momentarily zero in the vertical component; use this condition to identify the answer.
Derive an expression for the maximum height reached by an object thrown upward from the ground with an initial speed 𝑣0
Keywords: “derive,” “maximum height.”
– Answer Strategy: Set final velocity
𝑣=0 and use 0=v0^2-2𝑔 𝐻 to solve for
𝐻=𝑣0^2/2𝑔
.
Vector
Magnitude and Direction
Scalar
Magnitude
Function
How several quantities relate to one another
8,765 kg/m^3 to g/cm^3
8.777 g/cm^3
Instantaneous velocity
The velocity at a small time interval/ specific time
Instantaneous Acceleration
acceleration over a time period that small/ at a specific time
P v T Graph slope represents
Velocity
V v T Graph Slope represents
acceleration
V vT Graph area under the curve represents
Displacement
Area under the Curve of a Acceleration v T graph is
Change in velocity
Y component of a Vector typical 90 degree triangle
sin(0) x hypotenuse = Y component
The peak of a projectile velocity is
Zero
