Physics Chapter: Motion and Acceleration

<p>24 m/s (C)</p> Signup and view all the answers

<p>h = v_i t + rac{1}{2} g t^2 (C)</p> Signup and view all the answers

<p>It acts in opposition to the upward motion. (B)</p> Signup and view all the answers

<p>It decreases until it reaches the maximum height. (A)</p> Signup and view all the answers

<p>Total displacement can be calculated as the sum of ascent and descent. (C)</p> Signup and view all the answers

<p>Maximum height can be calculated using the time taken. (C)</p> Signup and view all the answers

<p>Just before hitting the ground. (C)</p> Signup and view all the answers

<p>The cyclist is moving at a constant speed. (B)</p> Signup and view all the answers

<p>The cyclist is maintaining a constant speed. (B)</p> Signup and view all the answers

<p>By slopes of the lines. (D)</p> Signup and view all the answers

<p>The cyclist travels 20 meters. (D)</p> Signup and view all the answers

<p>7 seconds (D)</p> Signup and view all the answers

<p>Changes in velocity (D)</p> Signup and view all the answers

<p>Use the area under the graph. (D)</p> Signup and view all the answers

<p>The cyclist is accelerating. (B)</p> Signup and view all the answers

<p>Changes in speed over time. (A)</p> Signup and view all the answers

<p>It creates a horizontal line. (D)</p> Signup and view all the answers

<p>3.0 m/s² (C)</p> Signup and view all the answers

<p>28.5 m/s (C)</p> Signup and view all the answers

<p>2.5 seconds (C)</p> Signup and view all the answers

<p>60 m (B)</p> Signup and view all the answers

<p>3.125 m/s² (D)</p> Signup and view all the answers

<p>Increases steadily (B)</p> Signup and view all the answers

<p>v = u + at (B)</p> Signup and view all the answers

<p>They serve as reference points for time calculations. (B)</p> Signup and view all the answers

<p>The train travels equal distances in equal time intervals. (D)</p> Signup and view all the answers

<p>24.5 m/s (A)</p> Signup and view all the answers

<p>25 m s–1 (B)</p> Signup and view all the answers

<p>10 seconds (A)</p> Signup and view all the answers

<p>t = 10 to t = 18 s (A)</p> Signup and view all the answers

<p>30 seconds (A)</p> Signup and view all the answers

<p>8 seconds (A)</p> Signup and view all the answers

<p>526 m (B)</p> Signup and view all the answers

<p>Total distance and time between points (B)</p> Signup and view all the answers

<p>Deceleration (A)</p> Signup and view all the answers

<p>It decreases (D)</p> Signup and view all the answers

<p>Two straight lines with one slope representing deceleration (D)</p> Signup and view all the answers

<p>Constant speed (C)</p> Signup and view all the answers

<p>The total distance covered (B)</p> Signup and view all the answers

<p>25 m/s (C)</p> Signup and view all the answers

<p>t = 20 s (A)</p> Signup and view all the answers

<p>40 m/s (C)</p> Signup and view all the answers

<p>25 m/s (A)</p> Signup and view all the answers

<p>10 seconds (C)</p> Signup and view all the answers

<p>2.5 m/s² (A)</p> Signup and view all the answers

<p>t = 20 s (C)</p> Signup and view all the answers

<p>10 m/s (C)</p> Signup and view all the answers

<p>t = 40 s (B)</p> Signup and view all the answers

<p>75 m (B)</p> Signup and view all the answers

<p>From t = 20 s to t = 30 s (A)</p> Signup and view all the answers

<p>10 seconds (A)</p> Signup and view all the answers

<p>Train Q (A)</p> Signup and view all the answers

<p>t = 30 s (D)</p> Signup and view all the answers

<p>23.5 m (D)</p> Signup and view all the answers

<p>v = \sqrt{u^2 + 2gh} (A)</p> Signup and view all the answers

<p>3.0 s (A)</p> Signup and view all the answers

<p>25.5 m (A)</p> Signup and view all the answers

<p>It decelerates until it stops and then accelerates downwards. (A)</p> Signup and view all the answers

<p>0 m/s (B)</p> Signup and view all the answers

<p>It reduces the ball's maximum height. (D)</p> Signup and view all the answers

<p>45.9 m (A)</p> Signup and view all the answers

<p>It would increase. (B)</p> Signup and view all the answers

<p>The same distance it travels upwards. (B)</p> Signup and view all the answers

<p>The weight of the ball. (D)</p> Signup and view all the answers

<p>Its speed is decreasing. (B)</p> Signup and view all the answers

<p>Sum of potential and kinetic energy (B)</p> Signup and view all the answers

The initial upward velocity of the stone.

The time taken for the stone to hit the ground from the point it was thrown.

The acceleration due to gravity, acting downwards on the stone.

The distance the stone travels downwards from its initial position to the ground.

The final velocity of the stone as it hits the ground.

A formula that relates displacement, initial velocity, final velocity, time, and acceleration.

The equation of motion that connects displacement, initial velocity, final velocity, time and acceleration.

The initial height from which the stone was thrown.

The final velocity of the stone just before it hits the ground.

An equation that connects initial velocity, final velocity, acceleration and time.

The rate of change of velocity over time. In this context, it's the constant acceleration that both trains experience initially.

The speed at which an object is moving. In this case, it's measured in meters per second (m/s).

The time elapsed since the start of the motion, measured in seconds.

The point at which Train Q overtakes Train P.

The time at which Train Q overtakes Train P, measured in seconds.

The type of motion where the speed remains constant, meaning the object is not accelerating.

The type of motion where the velocity changes at a constant rate. This means the speed is increasing or decreasing steadily.

A visual representation of the speed of an object over time, usually plotted with time on the horizontal axis and speed on the vertical axis.

The moment when both trains are traveling at the same speed. This means they have equal velocities at this specific point in time.

The distance traveled by each train, which will be represented by the area under their respective speed-time graphs.

A point on the speed-time graph where the train stops accelerating and maintains a constant speed.

The initial conditions of each train, including their starting positions and speeds. Both are at rest and level with each other.

Train P starts moving at time t = 0, while Train Q starts later at t = 20 seconds.

Train P reaches a constant speed of 25 m/s after accelerating for 10 seconds.

Train Q reaches a constant speed of 40 m/s after accelerating for an unknown period of time.

The rate at which the train's velocity changes over time.

The speed of the train when it passes point C.

The time taken for the train to travel from point B to point C.

The speed of the train when it passes point A.

The distance between points A and B.

The time taken for the train to travel from point A to point B.

The distance between points B and C.

The train's motion with a constant change in velocity.

The cyclist is traveling at a constant acceleration. This means the cyclist's speed is changing by the same amount every second.

The cyclist is traveling at a constant speed, meaning the speed is not changing.

The area under the speed-time graph represents the distance traveled.

The area of the triangular portion of the graph is calculated using 1/2 × base × height. The base of the triangle is 3 seconds (3 - 0), and the height is 5 m/s. So the area is 1/2 × 3 × 5 = 7.5 m.

The distance traveled during the next 4 seconds (from 3 to 7 seconds) is calculated using the area of a rectangle. The rectangle has a length of 4 s (7 - 3) and a width of 5 m/s. The area is calculated as 4 × 5 = 20 m

The total distance traveled is the sum of the distances traveled in each section: 7.5 m + 20 m = 27.5 m.

The initial speed at which the ball is projected upwards from point A.

The maximum height the ball reaches above its starting point A.

The speed the ball has just before it hits the ground.

The total time it takes for the ball to travel from point A to the ground.

The acceleration due to gravity, acting downwards on the ball.

The distance between point A and the ground.

This is the point in the ball's path where its upward velocity becomes zero before it starts falling back down.

The upward journey of the ball from its launch point to the highest point.

The downward journey of the ball from its highest point to reaching the ground.

The specific point 1.5 meters above the ground from where the ball is initially launched.

The upward velocity of the ball at the highest point.

The point on the ground where the ball finally hits.

The total distance travelled by the ball from point A to the ground, including both the upward and downward journey.

The specific time taken for the ball to travel from point A to the highest point.

The specific time taken for the ball to travel from the highest point to the ground.

The total distance traveled by the car from point A to point B. It's a crucial piece of information to calculate different parameters of the car's motion.

The speed of the car after it has decelerated for 8 seconds (at time t = 18 s). It's a key value to calculate the distance traveled during deceleration.

The time period between 10 s and 18 s when the car is undergoing uniform deceleration. This is the time interval where the car is slowing down steadily.

The time period between 10 s and 30 s when the car is either decelerating or moving at a constant speed. It's important to identify the specific time intervals for calculating total distance.

The speed maintained by the car before it begins to decelerate. It's constant during the first 10 seconds of the journey.

The period of time between t = 0 and t = 10 s, during which the car moves at a constant speed. It's another important time interval to analyze the car's overall motion.

The section of the speed-time graph that represents the car moving at a constant speed. It's a horizontal line as the speed is unchanging during this time.

The section of the speed-time graph that represents the car decelerating. It's a sloping line downwards as the speed decreases over time.

The point on the speed-time graph where the car begins to decelerate. It marks the transition between the constant speed section and the deceleration section of the graph.

The area under the speed-time graph between time t = 0 and t = 10 seconds. It's a straightforward rectangular shape, providing a direct way to calculate the distance traveled during that time.

The area under the speed-time graph between time t = 10 seconds and t = 18 seconds. This will be a triangular shape representing the distance traveled during deceleration.

The area under the speed-time graph between time t = 18 seconds and t = 30 seconds. It will be a rectangle representing the distance traveled at constant speed after deceleration.

The total area under the entire speed-time graph from t = 0 to t = 30 seconds. It represents the overall distance traveled by the car from point A to point B.