Kinematics, Dynamics, and Rotational Motion: Step-by-Step Physics Guidance

Q1. A particle starts at $x = -2$. From $t = 0$ to $t = 3$ s it accelerates at $2\ \mathrm{m/s^2}$. From $t = 3$ to $t = 6$ s it moves at a constant velocity. From $t = 6$ to $t = 8$ s it decelerates to a stop. What is its final position at $t = 8$ s?

Background

Topic: Kinematics (One-Dimensional Motion)

This question tests your understanding of motion with constant acceleration, constant velocity, and deceleration. You'll need to break the motion into segments and apply kinematic equations to each.

Key Terms and Formulas

• Displacement: $\Delta x = x_f - x_i$

• Constant acceleration: $x = x_0 + v_0 t + \frac{1}{2} a t^2$

• Final velocity: $v = v_0 + a t$

• Constant velocity: $x = x_0 + v t$

Step-by-Step Guidance

1. For $t = 0$ to $t = 3$ s, identify the initial position ($x_0 = -2$), initial velocity ($v_0 = 0$), and acceleration ($a = 2\ \mathrm{m/s^2}$). Use the kinematic equation to find the position and velocity at $t = 3$ s.

   $x_1 = x_0 + v_0 t_1 + \frac{1}{2} a t_1^2$

   $v_1 = v_0 + a t_1$

2. For $t = 3$ to $t = 6$ s, the particle moves at the constant velocity found at $t = 3$ s. Use the constant velocity equation to find the position at $t = 6$ s.

   $x_2 = x_1 + v_1 (t_2 - t_1)$

3. For $t = 6$ to $t = 8$ s, the particle decelerates to a stop. The initial velocity for this segment is the velocity at $t = 6$ s (which is the same as $v_1$). The final velocity is $0$. Use the kinematic equation to find the acceleration during this segment:

   $v_f = v_i + a_2 (t_3 - t_2)$

   Set $v_f = 0$ and solve for $a_2$.

4. Use the found acceleration to calculate the displacement during this segment:

   $x_3 = x_2 + v_i (t_3 - t_2) + \frac{1}{2} a_2 (t_3 - t_2)^2$

Try solving on your own before revealing the answer!

Q2. A car passes a stationary police cruiser at a constant $30\ \mathrm{m/s}$. The cruiser waits $1$s and then accelerates at a constant rate$a$. If the cruiser catches the car at the $300$ m mark, what was the cruiser's acceleration?

Background

Topic: Kinematics (Relative Motion, Constant Acceleration)

This problem involves two objects: one moving at constant velocity and one starting from rest after a delay and accelerating. You need to set up equations for both and solve for the acceleration when their positions are equal.

Key Terms and Formulas

• Constant velocity: $x = x_0 + v t$

• Constant acceleration from rest: $x = x_0 + v_0 t + \frac{1}{2} a t^2$

Step-by-Step Guidance

1. Write the position equation for the car: it starts at $x = 0$ and moves at $30\ \mathrm{m/s}$ for all $t$.

   $x_{car} = 30 t$

2. The cruiser starts from rest after $1$s, so its motion starts at$t = 1$s. For$t > 1$ s, its position is:

   $x_{cruiser} = \frac{1}{2} a (t - 1)^2$

3. Set the positions equal at the catch-up point ($x = 300$ m):

   $30 t = \frac{1}{2} a (t - 1)^2 = 300$

4. Solve for $t$ using the car's equation, then substitute into the cruiser's equation to solve for $a$.

Try solving on your own before revealing the answer!

Q3. Jupiter has a radius of $69,600$ km and rotates once every $9.93$ hrs. Calculate the tangential velocity and the centripetal acceleration of a person standing at a latitude of $65$ degrees.

Background

Topic: Rotational Motion, Circular Motion

This question tests your ability to relate rotational motion to linear (tangential) velocity and centripetal acceleration, accounting for latitude on a rotating sphere.

Key Terms and Formulas

• Angular velocity: $\omega = \frac{2\pi}{T}$

• Effective radius at latitude $\theta$: $r = R \cos(\theta)$

• Tangential velocity: $v = \omega r$

• Centripetal acceleration: $a_c = \frac{v^2}{r}$

Step-by-Step Guidance

1. Convert Jupiter's radius to meters and the rotation period to seconds.

2. Calculate the angular velocity $\omega$ using $T$.

   $\omega = \frac{2\pi}{T}$

3. Find the effective radius at $65^\circ$ latitude: $r = R \cos(65^\circ)$

4. Calculate the tangential velocity: $v = \omega r$

5. Set up the formula for centripetal acceleration: $a_c = \frac{v^2}{r}$

Try solving on your own before revealing the answer!

Q4. A fish swims in a river where the current flows due West at $16$mph. The fish's speed relative to the water is $22$ mph. The fish wants to move in a direction such that its resultant path (relative to the ground) is pointed directly South. At what angle must the fish aim its body relative to the water to compensate for the current? What is the fish’s actual speed relative to the ground $V_{SG}$? The "Heading" Challenge: Translate your angle into a navigational heading (e.g., “theta degrees East of South").

Background

Topic: Relative Velocity, Vector Addition

This problem involves vector addition to find the direction and magnitude of the fish's velocity relative to the ground, compensating for the river's current.

Key Terms and Formulas

• Relative velocity: $\vec{V}_{SG} = \vec{V}_{SW} + \vec{V}_{WG}$

• Trigonometry: $\sin, \cos, \tan$ for vector components

Step-by-Step Guidance

1. Draw a vector diagram showing the fish's velocity relative to the water ($22$mph at an angle$\theta$east of south) and the water's velocity relative to the ground ($16$ mph west).

2. Set up the vector components so that the resultant velocity points directly south. Write equations for the east-west and north-south components.

3. Set the west-east component of the resultant velocity to zero (since the fish wants to go directly south):

   $22 \sin \theta = 16$

4. Solve for $\theta$ and then use the southward component to find the fish's actual speed relative to the ground:

   $V_{SG} = 22 \cos \theta$

5. Translate $\theta$ into a navigational heading (e.g., degrees east of south).

Try solving on your own before revealing the answer!

Q5. A basketball is thrown from an initial height of $2$m toward a hoop $9.2$ m away that is $3.1$ m high. If the ball takes $1.6$ s to reach the hoop, find the initial velocity magnitude $v_i$ and the launch angle $\theta$.

Background

Topic: Projectile Motion

This question tests your ability to analyze projectile motion in two dimensions, using kinematic equations to relate time, displacement, and initial velocity components.

Key Terms and Formulas

• Horizontal motion: $x = v_{i,x} t$

• Vertical motion: $y = y_0 + v_{i,y} t - \frac{1}{2} g t^2$

• Initial velocity components: $v_{i,x} = v_i \cos \theta$, $v_{i,y} = v_i \sin \theta$

Step-by-Step Guidance

1. Write the horizontal displacement equation using the known distance and time:

   $9.2 = v_i \cos \theta \times 1.6$

2. Write the vertical displacement equation using the initial and final heights, time, and gravity:

   $3.1 = 2 + v_i \sin \theta \times 1.6 - \frac{1}{2} (9.8) (1.6)^2$

3. Solve the first equation for $v_i \cos \theta$ and substitute into the second equation to solve for $v_i \sin \theta$.

4. Use the two equations to solve for $v_i$ and $\theta$ (hint: use $\sin^2 \theta + \cos^2 \theta = 1$ if needed).

Try solving on your own before revealing the answer!