Question Bank

The lecture suggests that unit conversions are more crucial because they directly impact the ability to solve physics problems accurately and understand the relationships between different physical quantities. While significant figures are important in general scientific practice, the AP Physics 1 exam does not heavily emphasize them, allowing students to focus on the core physics concepts and problem-solving strategies.

The process involves multiplying the original quantity by conversion factors that are equal to one. In the example, kg/m³ is multiplied by (1000 g / 1 kg) and (1 m / 100 cm)³, where each fraction equals one. This allows for the cancellation of unwanted units and the introduction of desired units without changing the value of the original quantity.

The statement is incorrect. When converting from a larger unit to a smaller unit, the numerical value will increase because it takes more of the smaller unit to represent the same quantity. For example, 1 meter is equal to 100 centimeters; the numerical value increased from 1 to 100.

Distance is the total path length traveled, so the distance is 5m + 3m = 8m. Displacement is the change in position, a vector from the starting point to the ending point. The magnitude of the displacement is found using the Pythagorean theorem: sqrt(5^2 + 3^2) ≈ 5.83m.

Velocity is a vector quantity, meaning it has both magnitude (speed) and direction. An object can have constant speed but changing velocity if its direction is changing. For example, an object moving in a circle at a constant speed has a changing velocity because its direction is constantly changing.

First, establish a scale (e.g., 1 cm = 10 N). Then, draw an arrow starting from the origin. The length of the arrow should correspond to the magnitude of the force based on your scale (e.g., 5 cm). Finally, use a protractor to ensure the arrow is drawn at an angle of 30 degrees above the horizontal axis.

The runner's displacement is zero because their final position is the same as their initial position. The distance they ran is 400 meters. This difference highlights that displacement only considers the change in position, while distance considers the total path traveled, regardless of direction.

If a car travels 10 meters in a straight line without changing direction, the magnitude of its displacement is equal to the distance traveled (10 meters) because the path is a straight line. However, if the car travels 10 meters forward and then 5 meters backward, the distance traveled is 15 meters, while the magnitude of the displacement is only 5 meters because it is the straight line distance from start to finish.

Displacement is a vector quantity because it has both magnitude (the distance between the initial and final positions) and direction. For example, saying a hiker's displacement is 5 km is incomplete; we need to specify the direction (e.g., 5 km North) to fully describe their change in position. Without the direction, we don't know where the hiker ended up relative to their starting point.

Average speed is the total distance traveled divided by the time taken, while average velocity is the displacement (change in position) divided by the time taken. For example, if a car travels 10 km east and then 10 km west in one hour, its average speed is 20 km/h, but its average velocity is 0 km/h. It's important to distinguish them because velocity considers direction, which is crucial in many physics problems.

While instantaneous velocity is related to average velocity over a short time interval, it's not simply the average velocity. Instantaneous velocity is the limit of the average velocity as the time interval approaches zero. Mathematically, it's the derivative of the position function with respect to time, representing the rate of change of position at a specific instant.

The car has a constant average speed because the distance it travels around the circle increases linearly with time. However, the car's average velocity over a complete circle is zero because its displacement is zero (it returns to its starting point). Over a portion of the circle, the average velocity would be non-zero, but its direction would be constantly changing.

Average acceleration is the change in velocity over a time interval, while instantaneous acceleration is the acceleration at a specific moment. For example, in a car crash, the average acceleration over a few seconds might seem survivable, but the instantaneous acceleration at the moment of impact could be deadly.

This is incorrect because acceleration is the rate of change of velocity, not the velocity itself. An object can have zero velocity at a specific instant while still experiencing acceleration. For example, a ball thrown straight up has zero velocity at its highest point, but its acceleration due to gravity is still acting downwards.

The average acceleration is 0 m/s² because the velocity is constant, meaning there's no change in velocity over any time interval. Similarly, the instantaneous acceleration is also 0 m/s² at any point in time because the velocity isn't changing at that specific moment either.

The UAM equations are derived under the assumption of constant acceleration. Using the average acceleration in a situation with non-uniform acceleration will not accurately represent the motion at any specific instant. Instead, calculus-based methods (integration) are required to determine displacement and velocity when acceleration varies with time.

If we define upwards as positive, the initial velocity is positive. At the peak, the final velocity is momentarily zero. Acceleration due to gravity is always downwards, so it's negative. The displacement from the initial point to the peak is positive because the ball moved upwards.

We can use the equation: displacement = 1/2 * (v_f + v_i) * t. Since the car starts from rest, v_i = 0. Therefore, the displacement is 1/2 * v * t. This equation is suitable because it directly relates displacement to final velocity, initial velocity, and time, all of which are given.

The student's statement is incorrect. Acceleration is the rate of change of velocity, not the velocity itself. Even though the velocity is zero at the peak, the object is still under the influence of gravity, resulting in a constant downward acceleration of approximately 9.81 m/s² (or 10 m/s² for AP Physics).

Consider a ball thrown downwards from a cliff. If we define the downward direction as positive, the initial velocity should be positive. If we define upward as positive, the initial velocity is negative. Choosing the incorrect sign will lead to an incorrect calculation of the final velocity or the time it takes to reach the ground, as the equations rely on consistent directional representation.

Using the UAM equation d = v₀t + (1/2)at², since v₀ = 0, we have h = (1/2)at². Solving for t, we get t = sqrt(2h/a). If 'a' is doubled, the time 't' becomes t = sqrt(2h/2a) = sqrt(h/a), which is 1/sqrt(2) times the original time. Therefore, the time would decrease by a factor of sqrt(2).

The position vs. time graph would be a curve with increasing positive slope (representing increasing velocity). The velocity vs. time graph would be a straight line with a positive slope (representing constant positive acceleration). The acceleration vs. time graph would be a horizontal line above the x-axis (representing constant positive acceleration).

A velocity vs. time graph does not directly show the object's position at a given time. Instead, it shows the object's velocity at a given time. The area under the velocity vs. time curve represents the *change* in position (displacement) over a time interval, not the absolute position.

A car slowing down while moving forward has positive velocity and negative acceleration. The velocity vs. time graph would be a line with a negative slope, indicating that the velocity is decreasing over time. The negative slope represents the negative acceleration, meaning the object is decelerating.

Uniformly accelerated motion is characterized by a parabolic position-time graph, indicating a constant rate of change in velocity. Non-uniform acceleration would result in a position-time graph that is not a perfect parabola, potentially showing curves with varying concavity. For example, free fall near Earth's surface is uniformly accelerated, while a rocket launch with increasing thrust would exhibit non-uniform acceleration.

The statement is a misconception because the acceleration due to gravity depends on the planet's mass and radius, not just the act of throwing an object upwards. While -10 m/s² is a good approximation for Earth, the acceleration would be different on other planets with different gravitational fields. The acceleration is determined by the gravitational force acting on the object, which is proportional to the planet's mass and inversely proportional to the square of the distance from the planet's center.

To determine if the motion is free fall, measure the object's acceleration. If the acceleration is constant and downwards, it's likely free fall. To identify the planet, compare the measured acceleration to the known surface gravity values of different planets. If the acceleration matches a known value, that's a likely candidate.

Memorized rules are only valid for angles measured with respect to a specific axis (usually the horizontal). If the angle is measured with respect to a different axis (like the vertical, as in the example), the trigonometric relationships are reversed. Deriving the components using SOH CAH TOA ensures the correct relationship is applied based on the angle's orientation.

The x-component of the initial velocity (Vx) tells us the constant horizontal velocity of the projectile throughout its flight, assuming negligible air resistance. The y-component of the initial velocity (Vy) tells us the initial upward velocity, which is affected by gravity and determines the time it takes to reach the peak of its trajectory and the overall flight time.

The statement is incorrect because when the angle is 0 degrees with respect to the horizontal, the vector lies entirely along the x-axis. Therefore, the x-component would be 10 units (10*cos(0)), and the y-component would be 0 units (10*sin(0)).

The statement is partially correct because only the vertical component of velocity (Vy) is zero at the peak of the projectile's trajectory. The horizontal component of velocity (Vx) remains constant throughout the motion (assuming negligible air resistance) and is non-zero at the peak.

Since the launch and landing points are at the same height, the time to reach the maximum height is half of the total time of flight. Using the initial vertical velocity and the acceleration due to gravity, we can calculate the time to reach the maximum height using kinematic equations. Doubling this time gives the total time the projectile is in the air.

Separating the motion into horizontal and vertical components simplifies the problem because the acceleration only acts in the vertical direction (due to gravity). The horizontal motion has zero acceleration, making it constant velocity motion. The key assumption is that air resistance is negligible, allowing us to treat the horizontal and vertical motions as independent.

Your velocity relative to the person outside the train is 22 m/s East. This is because the observer sees both the train's velocity and your walking velocity adding together. This illustrates how different frames of reference (the train vs. the ground) result in different observed velocities.

The boat's velocity relative to the riverbank is the vector sum of its velocity relative to the water and the water's velocity. The magnitude is found using the Pythagorean theorem (√(5^2 + 3^2) ≈ 5.83 m/s), and the direction is North-East at an angle of arctan(3/5) relative to North.

While the difference in speeds gives the magnitude of the relative velocity in this specific scenario (same direction), it's not universally true. Relative velocity is a vector quantity, and direction matters. If the cars were moving in opposite directions, you would add the speeds to find the magnitude of the relative velocity.