DP IB Physics HL: Kinematics

Comprehensive review flashcards covering Kinematics for DP IB Physics HL, including distance, displacement, speed, velocity, acceleration, SUVAT equations, motion graphs, projectile motion, fluid resistance, and terminal speed.

What is the difference between distance and displacement in terms of quantity type and definition?

Distance is a scalar quantity measuring the total path length an object travels without regard to direction. Displacement is a vector quantity measuring the change in position from the starting point to the final position, describing both magnitude and direction.

A professor walks around her rectangular garden of dimensions $15\,\text{km}$ by $9\,\text{km}$ following the path ABCDA. What are her total distance travelled and displacement upon returning to point A?

The total distance travelled is $15 + 9 + 15 + 9 = 48\,\text{km}$. The displacement is $0\,\text{km}$ because she returns to her starting position.

How is velocity defined, and how does it differ fundamentally from speed?

Velocity is defined as the rate of change of displacement and is a vector quantity having both magnitude and direction. Speed is a scalar quantity that only possesses magnitude and represents the distance travelled per unit time.

How is instantaneous velocity determined from a curved displacement-time graph?

Instantaneous velocity is found by drawing a tangent to the curve at the specified time and calculating the gradient of that tangent.

Florence Griffith Joyner ran the women's $100\,\text{m}$ world record in 1988 with a time of $10.49\,\text{s}$. What was her average speed?

$$\text{average speed} = \frac{\text{total distance}}{\text{time taken}} = \frac{100\,\text{m}}{10.49\,\text{s}} = 9.53\,\text{m s}^{-1}$$

What is acceleration, and what are its standard SI units?

Acceleration is defined as the rate of change of velocity ($a = \frac{\Delta v}{\Delta t}$). It is a vector quantity with SI units of metres per second squared ($\text{m s}^{-2}$).

A bullet train decelerates uniformly in a straight line from an initial velocity of $50\,\text{m s}^{-1}$ to $42\,\text{m s}^{-1}$ over $30\,\text{s}$. What are the change in velocity and the acceleration of the train?

The change in velocity is $\Delta v = 42 - 50 = -8\,\text{m s}^{-1}$. The acceleration is $a = \frac{-8\,\text{m s}^{-1}}{30\,\text{s}} = -0.27\,\text{m s}^{-2}$, where the negative sign indicates deceleration.

What are the four kinematic (SUVAT) equations for an object moving with uniform acceleration?

1. $v = u + at$
2. $s = \frac{(u + v)}{2}t$
3. $s = ut + \frac{1}{2}at^2$
4. $v^2 = u^2 + 2as$

In kinematic problems, what default values are deduced from the phrases "starts from rest" and "falling due to gravity" (with downwards taken as positive)?

"Starts from rest" implies $u = 0\,\text{m s}^{-1}$ and $t = 0\,\text{s}$. "Falling due to gravity" implies an acceleration $a = g = 9.81\,\text{m s}^{-2}$.

A train travelling at $50\,\text{m s}^{-1}$ passes marker 1 and decelerates uniformly to $10\,\text{m s}^{-1}$ at marker 2 over a time interval of $20\,\text{s}$. What is the distance between the two markers?

$$s = \frac{(u + v)}{2}t = \frac{(50 + 10)}{2} \times 20 = 30 \times 20 = 600\,\text{m}$$

A cyclist passes friend B while moving at a constant velocity of $18\,\text{m s}^{-1}$. At that moment, B accelerates from rest at $1.5\,\text{m s}^{-2}$. How long does it take for B to catch up to cyclist A?

Equating displacements: $s_A = 18t$ and $s_B = \frac{1}{2}(1.5)t^2 = \frac{3}{4}t^2$. Setting $18t = \frac{3}{4}t^2$ yields $t^2 - 24t = 0$, so $t = 24\,\text{s}$.

A feather falls from rest in a vertical glass vacuum tube with a fall time of $0.5\,\text{s}$. Using $g = 9.81\,\text{m s}^{-2}$, what is the length $L$ of the tube?

$$L = ut + \frac{1}{2}gt^2 = 0 + \frac{1}{2} \times 9.81 \times (0.5)^2 = 1.2\,\text{m}$$

What do the slope and area under the curve represent on a displacement-time graph?

The slope equals velocity. The area under the curve is meaningless.

What do the slope and area under the curve represent on a velocity-time graph?

The slope equals acceleration. The area under the curve equals the change in displacement.

What do the slope and area under the curve represent on an acceleration-time graph?

The slope is meaningless. The area under the curve equals the change in velocity.

In which labelled section (A, B, C, D, or E) of this velocity-time graph is the cyclist's acceleration the largest, and why?

Section D, because it has the steepest slope (gradient) on the velocity-time graph.

Using this velocity-time graph of a vehicle starting from rest, what is the displacement of the vehicle at $40\,\text{s}$?

Displacement is the area of the triangle: $\text{Displacement} = \frac{1}{2} \times \text{base} \times \text{height}$. Converting $105\,\text{km h}^{-1}$ gives $\frac{105}{3600} = 0.0292\,\text{km s}^{-1}$. Thus, $\text{Displacement} = \frac{1}{2} \times 40 \times 0.0292 = 0.6\,\text{km} = 600\,\text{m}$.

For a ball bouncing vertically under gravity (upwards taken as positive), what are its velocity and acceleration at the highest point of the bounce?

At the highest point, its velocity is momentarily zero ($v = 0\,\text{m s}^{-1}$), and its acceleration is constant and directed downwards ($a = -g = -9.81\,\text{m s}^{-2}$).

How are the initial horizontal and vertical velocity components resolved for a projectile launched with speed $u$ at an angle $\theta$ to the horizontal?

The horizontal component is $u_x = u\cos(\theta)$ and the vertical component is $u_y = u\sin(\theta)$.

What is the formula for the maximum height $H$ attained by a projectile launched with initial velocity $u$ at an angle $\theta$ to the horizontal?

$$H = \frac{(u\sin(\theta))^2}{2g}$$

What is the formula for the horizontal range $R$ of a projectile launched on level ground with initial velocity $u$ at angle $\theta$?

$$R = \frac{u^2\sin(2\theta)}{g}$$

A stunt rider takes off horizontally from a height of $1.25\,\text{m}$ and lands $10\,\text{m}$ away. Ignoring air resistance, what was the initial take-off speed?

Vertical fall time: $t = \sqrt{\frac{2s}{g}} = \sqrt{\frac{2 \times 1.25}{9.81}} = 0.5\,\text{s}$. Horizontal speed: $v = \frac{s}{t} = \frac{10}{0.5} = 20\,\text{m s}^{-1}$.

What four characteristics define frictional fluid resistance (viscous drag)?

1. It always acts in the opposite direction to the motion of the object.
2. It never speeds an object up or initiates movement.
3. It always slows down an object or maintains constant speed.
4. It always transfers energy away from the object to the surroundings.

How does air resistance alter the flight time, range, maximum height, and trajectory shape of a projectile compared to a vacuum?

Air resistance decreases the flight time, decreases the horizontal range, decreases the maximum height, and distorts the trajectory so it is no longer a symmetrical parabola (it becomes steeper on the way down than on the way up).

What is terminal velocity, and what force condition occurs when a falling object attains it?

Terminal velocity is the constant maximum velocity reached by a falling object when the upward viscous drag force equals the downward weight, resulting in zero net force and zero acceleration.

Two skydivers, A and B, have identical surface area and volume, but skydiver A is heavier. Who will reach a higher terminal velocity, and who should jump first to reach terminal velocity at the same time?

Skydiver A will reach a higher terminal velocity because greater weight requires a larger drag force to balance. Skydiver B should jump first because skydiver B takes longer to reach terminal velocity.