Question 1: A velocity-time graph shows a straight line sloping upwards from 0 m/s at 0 seconds to 20 m/s at 10 seconds. Calculate the acceleration of the object.

Answer: 2 m/s^2

Question 2: If a velocity-time graph is horizontal, what does this indicate about the acceleration?

Answer: Zero or no acceleration

Question 3: A student incorrectly states that the slope of a velocity-time graph is the acceleration. Identify a common mistake that might lead to this misconception.

Answer: Confusing slope with velocity or misunderstanding that slope indicates acceleration

Question 1: A car accelerates uniformly from 5 m/s to 25 m/s over 8 seconds. Draw the velocity-time graph and calculate the acceleration. Explain any assumptions made.

Answer: Graph: straight line from (0,5) to (8,25). Acceleration = (25-5)/8 = 2.5 m/s^2

Question 2: Explain why a decreasing slope on a velocity-time graph indicates negative acceleration. Illustrate with an example scenario.

Answer: A decreasing slope means velocity is decreasing over time, indicating negative acceleration, e.g., a car braking

Question 3: A velocity-time graph shows a curve that starts steep and levels off. What does this suggest about the acceleration? Justify your answer.

Answer: The acceleration is decreasing, approaching zero as the curve levels off

Question 1: A cyclist accelerates from 0 m/s to 10 m/s over 5 seconds. The cyclist then maintains this speed. Sketch the velocity-time graph and describe the acceleration phases.

Answer: Initial slope up (positive acceleration), then horizontal line (zero acceleration).

Question 2: A train slows down from 30 m/s to 0 m/s in 15 seconds. Calculate the magnitude of the acceleration and discuss how this would look on a velocity-time graph.

Answer: Acceleration = (0 - 30)/15 = -2 m/s^2; graph shows a downward slope.

Question 1: An object moves along a straight line with velocity v(t) = 3t^2 - 4t + 2. Find the acceleration function and determine when the acceleration is zero. Show all steps.

Answer: a(t) = 6t - 4; set to zero: 6t - 4 = 0 => t = 2/3 sec.

Question 2: Construct a velocity-time graph for an object that accelerates for 5 seconds, then moves at constant speed, then decelerates. Describe the key features of your graph.

Answer: Graph starts with positive slope, then horizontal, then negative slope; representing acceleration, constant velocity, and deceleration.

Question 1: A student claims that the slope of a velocity-time graph is the displacement. Identify and correct this misconception.

Answer: Incorrect: The slope represents acceleration. Displacement is the area under the graph.

Question 2: A graph shows a curved line sloping upwards. The student interprets this as increasing acceleration. Is this correct? Explain and justify.

Answer: Not necessarily; a curve indicates changing acceleration. The slope shows instantaneous velocity, but the curvature indicates acceleration change.

Question 3: Identify a common mistake when calculating acceleration from a velocity-time graph and explain how to avoid it.

Answer: Mistake: Using the change in velocity over the change in time without considering the correct interval. To avoid, ensure correct start and end points.