AP Physics C Mechanics - 1.2 Displacement, Velocity, and Acceleration- Exam Style questions- FRQs
Displacement, Velocity, and Acceleration AP Physics C Mechanics FRQ
Unit: 1. Kinematics
Weightage : 10-15%
Question
_____ Increase _____ Decrease _____ Remain the same
ii. In a brief statement, describe the direction of the object’s acceleration and how the magnitude of this acceleration changed as the object fell.
iii. Using the graph, calculate an approximate value for the magnitude of the acceleration of the object at \(t = 0.20\,\text{s}\).
ii. Derive an expression for the magnitude of the net force \(F(t)\) exerted on the object as it falls through the fluid as a function of time \(t\).
ii. Determine the force exerted by the fluid on the object at this time. Justify your answer.
Most-appropriate topic codes (AP Physics C: Mechanics):
• Topic \(1.3\) — Representing Motion (Part \( \mathrm{a} \), \( \mathrm{b} \))
• Topic \(2.5\) — Newton’s Second Law (Part \( \mathrm{b} \), \( \mathrm{c} \))
• Topic \(2.9\) — Resistive Forces (Part \( \mathrm{b} \), \( \mathrm{c} \))
▶️ Answer/Explanation
(a)(i)
The correct choice is Increase.
Looking directly at the provided graph, the velocity values are getting larger (more positive) as time progresses.
This shows the object is continuing to speed up as it falls.
(a)(ii)
The object’s acceleration is directed downward because it is moving downward and speeding up.
However, the magnitude of the acceleration is decreasing as the object falls.
We can verify this because the slope of the velocity-time graph, which represents the acceleration, is getting flatter over time.
(a)(iii)
Acceleration is the slope of the tangent line on the velocity-time graph at \(t = 0.20\,\text{s}\).
Drawing a tangent line at this point and picking two points on that line (e.g., \((0.136\,\text{s}, 0.6\,\text{m/s})\) and \((0.226\,\text{s}, 0.8\,\text{m/s})\)) gives us:
\(a = \frac{\Delta v}{\Delta t}\)
\(a \approx \frac{0.8\,\text{m/s} – 0.6\,\text{m/s}}{0.226\,\text{s} – 0.136\,\text{s}}\)
\(a \approx 2.22\,\text{m/s}^2\)
(Any calculated slope between approximately 1.5 and 2.5 is acceptable depending on the drawn tangent line.)
(b)(i)
To find the vertical displacement, we need to integrate the velocity function with respect to time.
\(\Delta y = \int_{0}^{t} v(t’) \,dt’\)
\(\Delta y = \int_{0}^{t} A(1 – e^{-Bt’}) \,dt’\)
\(\Delta y = A \left[ t’ + \frac{1}{B}e^{-Bt’} \right]_{0}^{t}\)
\(\Delta y = A \left( t + \frac{1}{B}e^{-Bt} – \frac{1}{B} \right)\)
Substituting the known coefficients gives: \(\Delta y = 1.18 \left( t + \frac{1}{5}(e^{-5t} – 1) \right)\).
(b)(ii)
First, find the acceleration by taking the derivative of the velocity equation.
\(a(t) = \frac{dv}{dt} = \frac{d}{dt} \left[ A(1 – e^{-Bt}) \right]\)
\(a(t) = A B e^{-Bt}\)
Next, apply Newton’s second law (\(F = ma\)) and multiply by the given mass (\(12\,\text{g} = 0.012\,\text{kg}\)).
\(F(t) = m A B e^{-Bt}\)
\(F(t) = (0.012\,\text{kg})(1.18\,\text{m/s})(5\,\text{s}^{-1}) e^{-5t}\)
\(F(t) \approx 0.071 e^{-5t}\,\text{N}\).
(c)(i)
The constant (terminal) speed of the object is \(v = 1.18\,\text{m/s}\).
We know this because constant speed occurs after a long period of time.
If we take the limit as \(t \to \infty\), the term \(e^{-5t}\) approaches zero.
This leaves \(v = A = 1.18\,\text{m/s}\).
(c)(ii)
The force exerted by the fluid is approximately \(0.12\,\text{N}\).
When the object moves at a constant speed, its acceleration is zero, which means the net force is zero.
Therefore, the upward resistive force of the fluid perfectly balances the downward pull of gravity (weight).
\(F_{\text{fluid}} = mg\)
\(F_{\text{fluid}} = (0.012\,\text{kg})(10\,\text{m/s}^2)\)
\(F_{\text{fluid}} = 0.12\,\text{N}\).