SPH3U Midterm Motion Concepts Quiz

To find the acceleration of the car, we use the formula for uniform acceleration: \( a = \frac{Δv}{Δt} \), where \( Δv \) is the change in velocity and \( Δt \) is the time interval. Here, the car starts from rest (initial velocity \( u = 0 \)) and reaches a final velocity \( v = 24 \, \text{m/s} \) in \( t = 8 \, \text{s} \). Thus, the change in velocity \( Δv = v - u = 24 - 0 = 24 \, \text{m/s} \). Plugging these values into the formula gives \( a = \frac{24}{8} = 3 \, \text{m/s}² \).

Uniform motion is characterized by an object moving at a constant velocity, meaning both speed and direction remain unchanged over time. In contrast, non-uniform motion involves variations in either speed or direction, leading to changes in velocity. This distinction is crucial in physics, as it helps in understanding how objects behave under different conditions of motion.

On a velocity–time graph, the area under the curve represents the total displacement of an object over a given time period. This is because displacement is calculated by integrating velocity over time. When the velocity is plotted against time, the area between the curve and the time axis indicates how far the object has moved from its initial position, taking into account both the direction and magnitude of the velocity. Thus, the area directly correlates to the object's overall change in position.

To determine the time it takes for the ball to reach its maximum height, we can use the formula \( t = \frac{v}{g} \), where \( v \) is the initial velocity and \( g \) is the acceleration due to gravity. Here, the initial velocity is 20 m/s, and the downward acceleration due to gravity is 10 m/s². Plugging in the values, we get \( t = \frac{20 \, \text{m/s}}{10 \, \text{m/s²}} = 2 \, \text{s} \). This means it takes 2 seconds for the ball to reach its peak height before it starts to fall back down.

To find the horizontal displacement of the projectile, we use the formula: displacement = velocity × time. The horizontal velocity is 15 m/s, and the time of flight is 3 seconds. Therefore, the horizontal displacement is calculated as 15 m/s × 3 s = 45 m. This indicates that while the projectile falls vertically due to gravity, it also moves horizontally at a constant speed, resulting in a total horizontal distance of 45 meters before hitting the ground.

A straight horizontal line on a position–time graph indicates that the position of the object remains constant over time. This means that there is no change in the object's position, which signifies that it is not moving. Therefore, the object is at rest or stationary, as its velocity is zero, and it does not cover any distance regardless of the time elapsed.

A projectile launched horizontally starts its motion with an initial horizontal velocity and no vertical component. At the moment of launch, its vertical velocity is indeed zero because it has not yet begun to fall due to gravity. As it travels horizontally, gravity will then act on it, causing it to accelerate downward. Thus, the statement accurately reflects the initial conditions of a horizontally launched projectile.