University Of The District Of Columbia University Physics 20 ✓ Solved

Posted on December 16, 2025

University Of The District Of Columbiauniversity Physics 201homework I

Analyze and solve physics problems related to motion, vectors, displacement, velocity, acceleration, and related concepts, based on given experimental data and vector representations, including graphing and calculation of specific parameters at designated times.

Sample Paper For Above instruction

Introduction

The study of kinematics provides fundamental insights into the motion of objects without considering the causes of motion. This paper addresses several core topics within kinematics, including analyzing particle motion along a path, vector addition, displacement calculations, and motion under gravity. These concepts are applicable to a variety of real-world scenarios ranging from celestial body trajectories to vehicle motion analysis.

1. Motion Analysis of a Particle on the Moon

The primary focus begins with the analysis of a particle moving along the y-axis on the Moon, where its position varies with time according to the equation $ Y(t) = -50 + 100t - 20t^2 $. This quadratic function describes projectile-like motion influenced by lunar gravity.

Initial Position, Velocity, and Acceleration

The initial position $ Y(0) $ is computed by substituting $ t = 0 $:

Y(0) = -50 + 0 - 0 = -50\, \text{m}

The initial velocity $ v(t) $ is derived by differentiating $ Y(t) $:

v(t) = \frac{dY}{dt} = 100 - 40t

At $ t=0 $:

v(0) = 100\, \text{m/s}

Similarly, the acceleration $ a(t) $ is the second derivative of position or the derivative of velocity:

a(t) = \frac{dv}{dt} = -40\, \text{m/s}^2

which is constant throughout the motion.

Graphing Position and Velocity

The position $ Y(t) $ versus time is a downward-opening parabola with initial position at $-50\, \text{m}$ and initial velocity $100\, \text{m/s}$. The velocity versus time graph is a straight line with slope $-40\, \text{m/s}^2$. These graphs illustrate how the particle moves upward initially then decelerates under lunar gravity until reaching a maximum height, then descending.

Position, Velocity, and Acceleration at $t=10\,s$

Position:

Y(10) = -50 + 100(10) - 20(10)^2 = -50 + 1000 - 2000 = -1050\, \text{m}

Velocity:

v(10) = 100 - 40(10) = 100 - 400 = -300\, \text{m/s}

Acceleration:

a(t) = -40\, \text{m/s}^2 \quad \text{(constant)}

The negative velocity indicates the particle is moving downward at this point.

Maximum Height and Time

The maximum height occurs when velocity $ v(t) = 0 $:

0 = 100 - 40t \Rightarrow t = \frac{100}{40} = 2.5\, \text{s}

Position at this time:

Y(2.5) = -50 + 100(2.5) - 20(2.5)^2 = -50 + 250 - 125 = 75\, \text{m}

Thus, the maximum height reached by the particle is 75 meters, occurring at 2.5 seconds.

2. Vector Addition and Possibility of Summing to Zero

Considering vectors $ A = 2 $, $ B = 3 $, and $ C = 4 $, the question is whether their sum can be zero.

If these are scalar quantities, their algebraic sum is:

A + B + C = 2 + 3 + 4 = 9 \neq 0

which is non-zero unless considering vector components with specific directions.

If interpreted as vectors, for example, in a two-dimensional plane:

\vec{A} = 2 \hat{i}, \quad \vec{B} = 3 \hat{j}, \quad \text{and} \quad \vec{C} = 4 \text{ (assuming scalar magnitude along some axes)}

then, to determine if their vector sum can be zero, we analyze their components and directions. Generally, it's possible for the sum of vectors to be zero if they are appropriately directed, for example, $\vec{A} + \vec{B} + \vec{C} = 0$. The specific directions of these vectors determine if such a cancellation is physically achievable.

3. Vector Subtraction and Magnitude Calculation

Given vectors:

\vec{N} = 10 \hat{i} - 6 \hat{j}

\vec{M} = 5 \hat{i} - 8 \hat{j}

The vector $ \vec{L} = \vec{N} - \vec{M} $ is computed component-wise:

\vec{L} = (10 - 5) \hat{i} + (-6 + 8) \hat{j} = 5 \hat{i} + 2 \hat{j}

Magnitude of $ \vec{L} $:

|\vec{L}| = \sqrt{(5)^2 + (2)^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39\, \text{units}

Direction of $ \vec{L} $:

\theta = \arctan \left( \frac{2}{5} \right) \approx 21.8^\circ

with respect to the positive $ x $-axis.

4. Resultant Displacement from Multiple Vectors

Given three displacement vectors with magnitudes:

A = 20\, \text{m}, \quad B=15\, \text{m}, \quad C=20\, \text{m}

and assuming their directions are given, or executing calculations based on the vector components:

(a) Resultant in Unit-Vector Notation

Suppose vectors are oriented at specified angles (not provided here), we sum their components vectorially. For simplicity, assuming:

\vec{A} = 20 \hat{i}

\vec{B} = 15 \hat{j}

\vec{C} = 20 (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) \approx 20 (0.707 \hat{i} + 0.707 \hat{j}) = 14.14 \hat{i} + 14.14 \hat{j}

Sum:

\vec{R} = (20 + 14.14) \hat{i} + (15 + 14.14) \hat{j} = 34.14 \hat{i} + 29.14 \hat{j}

(b) Magnitude and Direction:

|\vec{R}| = \sqrt{(34.14)^2 + (29.14)^2} \approx \sqrt{1166 + 849} \approx \sqrt{2015} \approx 44.9\, \text{m}

Direction:

\theta = \arctan \left( \frac{29.14}{34.14} \right) \approx 40^\circ

from the positive x-axis.

5. Displacement of a Car from Velocity-Time Data

Using the velocity versus time graph (not provided numerically), the displacement is found by integrating the velocity over the specified time intervals:

(a) Between $t=0 \text{ and } 5\, s$

Approximate area under the curve or calculate using average velocity if data points are given. Assuming average velocity $ v_{avg} $ during this interval, then:

\text{Displacement} = v_{avg} \times 5\, s

Similarly for other intervals. Precise calculation requires velocity data.

(b) Between $t=13\, s$ and $t=15\, s$

Same method applies, integrating velocity over this interval.

(c) Distance traveled after 15 s

Total displacement from $ t=0 $ to $ t=15\, s $ is the sum of areas under the velocity-time graph, which absorbs past data.

6. Deceleration of a Car Coming to Rest

Given that the car takes 5 seconds to stop after traveling 200 meters:

\text{Initial velocity } v_0, \quad \text{acceleration } a

Using the kinematic equation:

v^2 = v_0^2 + 2 a s

At rest:

0 = v_0^2 + 2 a (200)

and time to halt:

v = v_0 + a t

Since $v=0$:

0 = v_0 + a \times 5

a = - \frac{v_0}{5}

Substituting into the first equation:

0 = v_0^2 + 2 \times (-v_0/5) \times 200

0 = v_0^2 - \frac{2 \times 200 v_0}{5} = v_0^2 - 80 v_0

Dividing through by $ v_0 $:

v_0 - 80 = 0 \Rightarrow v_0 = 80\, \text{m/s}

and acceleration:

a = - \frac{80}{5} = -16\, \text{m/s}^2

7. Two Balls Released at Different Conditions and Their Intersection Point

One ball is thrown upwards with initial velocity $ 50\, \text{m/s} $, and the other is released from a 6 m height from rest. Equations of motion:

Ball 1 (thrown upward):

y_1(t) = v_{0} t - \frac{1}{2} g t^2

with $ v_0 = 50\, \text{m/s} $.

Ball 2 (from the roof):

y_2(t) = 6 + 0 \times t - \frac{1}{2} g t^2 = 6 - 4.9 t^2

At the moment they pass each other:

y_1(t) = y_2(t)

50 t - 4.9 t^2 = 6 - 4.9 t^2

Simplify:

50 t = 6

t = \frac{6}{50} = 0.12\, \text{s}

Position:

y = 50 \times 0.12 - 4.9 \times (0.12)^2 \approx 6 - 0.07 \approx 5.93\, \text{m}

The balls pass each other approximately 0.12 seconds after release at a height of approximately 5.93 meters.

Conclusion

This collection of problems illustrates core principles of kinematics, vector algebra, and motion analysis. Applying derivatives, integrals, and vector component methods allows comprehensive understanding of particle dynamics and motion under various forces. These fundamental techniques underpin many engineering and physics applications extending from celestial mechanics to vehicular safety analysis.

References

Halliday, D., Resnick, R., & Walker, J. (2014). Fundamentals of Physics (10th ed.). Wiley.
Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (9th ed.). Cengage Learning.
Tipler, P. A., & Mosca, G. (2008). Physics for Scientists and Engineers (6th ed.). W. H. Freeman.
Young, H. D., & Freedman, R. A. (2012). University Physics (13th ed.). Pearson.
Knight, R. D., Jones, B., & Field, S. (2013). College Physics (8th ed.). Pearson.
Hibbeler, R. C. (2016). Engineering Mechanics: Dynamics (14th ed.). Pearson.
Tipler, P. A., & Mosca, G. (2017). Physics (7th ed.). W. H. Freeman.
Sears, F. W., Zemansky, M. W., & Young, H. D. (2014). University Physics with Modern Physics (13th ed.). Pearson.
Cutnell, J. D., & Johnson, K. W. (2012). Physics (9th ed.). Wiley.
Hassani, S. H. (2008). Vector and Tensor Analysis with Applications. Springer.

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