Read the Motion
A car is moving to the right but slowing down. Which statement must be true?
A. Its velocity and acceleration are both positive.
B. Its velocity is positive and its acceleration is negative.
C. Its velocity is negative and its acceleration is positive.
D. Its acceleration is zero because it is slowing down.
Answer: B
Explanation: Velocity tells us the direction of motion. Acceleration describes how velocity changes. The car moves right, so its velocity is positive, but it is slowing down, so acceleration is negative.
What happens at the top?
Multiple choice
Question
A ball is thrown vertically upward. At the highest point:
A. Velocity and acceleration are both zero.
B. Velocity is zero; acceleration is downward.
C. Acceleration changes from upward to downward.
D. Velocity is upward; acceleration is zero.
Answer: B. The ball's velocity is momentarily zero, but gravity still accelerates it downward.
An airplane flies horizontally at a constant speed of 100 m/s, at an altitude of 500 m.
It releases a package without changing its speed or altitude.
Ignore air resistance.
At the instant the package hits the ground, what is the straight-line distance between the airplane and the package?
A. 0 m
B. 500 m
C. 1,000 m
D. Approximately 1,100 m
Answer: B
Acceleration always means speeding up?
A student says:
"If an object is accelerating, it must be speeding up."
Is this always true? Explain.
Answer: No. Acceleration means velocity is changing, which can mean speeding up, slowing down, or changing direction.
Explanation: An object moving right with acceleration left will slow down. Whether speed increases depends on the relationship between velocity and acceleration.
"An object can be moving even when its acceleration is zero."
Convince me I'm right with an example, or prove me wrong.
Answer: True. An object moving at constant nonzero velocity has zero acceleration because its velocity isn't changing.
At one instant, an object's acceleration is zero. What can we conclude?
A. It must be stopped.
B. It must have constant velocity throughout the entire motion.
C. It could still be moving, even if its velocity is not changing at that instant.
D. It must be reversing direction
Answer: C
An object is thrown straight up into the air, if we ignore air resistance. then which is greater?
The time for
i) ascending
ii) descending
iii) the same
answer: iii the same
why: gravity acts the same on the object in both directions and due to the symmetry of the problem they are the same.
Two Directions, One Motion
300 points
Two identical marbles leave the same table at the same instant.
Marble A rolls horizontally off the edge. Marble B is simply dropped from rest.
Ignoring air resistance, which statements are true?
I. Both marbles have the same acceleration.
II. Both marbles hit the ground at the same time.
III. Both marbles have the same velocity just before impact.
Answer: I and II only.
Why: Both marbles experience the same gravitational acceleration, and their initial vertical velocities are identical, so they land simultaneously. However, Marble A retains horizontal velocity, so their total velocity vectors at impact differ.
A student sees a horizontal line at +5 m/s on a velocity–time graph and says:
"The object isn't moving because the graph is flat."
What's wrong with this interpretation?
Answer: A horizontal line on a velocity–time graph means constant velocity, not zero velocity. The object is moving at +5 m/s. Meaning what?
Explanation: The graph's height indicates velocity; its slope indicates acceleration. The area under tells us the displacement. Zero slope means zero acceleration.
A student looks at the velocity–time graph below and says:
"The line is flat, so the object must be stopped!"
The line is flat at 5m/s
What's wrong with their reasoning?
A) Nothing! A flat line means the object is not moving.
B) The object is moving at a constant velocity of +5 m/s, with zero acceleration.
C) The object is accelerating at a constant rate of +5 m/s².
D) The object is moving forward but gradually slowing down.
Correct answer: B
The graph shows a constant velocity of +5 m/s, meaning the object continues moving in the positive direction without speeding up or slowing down.
Height of the graph: Velocity (+5 m/s).
Slope of the graph: Acceleration (0 m/s²).
Follow-up challenge: What if the exact same horizontal line appeared on a position–time graph instead?
Key takeaway: Always check the axes before interpreting a graph. A flat line means something is constant, but what is constant depends on the vertical axis.
A student walks 6.0 m east, then 8.0 m north, and finally 3.0 m west.
Which statement correctly describes the student's resultant displacement?
A) 17.0 m, because displacement is the sum of all three movements.
B) 8.5 m, approximately 69° north of east, because the east–west movements partially cancel.
C) 10.0 m, approximately 53° north of east, because only the first two movements determine the displacement.
D) 8.5 m, approximately 21° north of east, because the net horizontal displacement is 3.0 m.
Correct answer: B
B is reasonable bigger vertical --> bigger angle n of e
Step 1: Combine horizontal components.
Step 2: Identify vertical displacement.
Step 3: Calculate the resultant displacement.
Step 4: Find the direction.
A ball is thrown straight upward at 19.6 m/s. You want to know how high it rises above its launch point.
You could calculate the time to the top first, but is there a more direct method?
Choose a relationship and explain why.
Answer: Yes:
vfy^2=viy^2+2ay
At the top, vertical velocity is zero, so:
0=(19.6)^2+2(-9.8)y
y=19.6
Explanation: This relationship uses initial velocity, final velocity, and acceleration to find displacement without first calculating time.
A ball rolls horizontally off a 4.9 m high platform with a speed of 3.0 m/s.
Which approach would correctly determine how far the ball travels horizontally?
A) Analyze horizontal motion first to determine the time in the air, then use vertical motion to find the range.
B) Analyze vertical motion first to determine the time in the air, then use horizontal velocity to find the range.
C) Combine the horizontal and vertical velocities using the Pythagorean theorem, then use the resulting speed to calculate the range.
D) Use the platform's height and horizontal velocity directly in a kinematics equation to calculate the range
Correct Answer: B
Why?
Vertical motion tells us how long the ball is in the air. Horizontal motion tells us how far it travels during that time.
Step 1: Vertical motion
t
Step 2: Horizontal motion
x
Key takeaway: The horizontal and vertical motions are independent, but they share the same time.
Jeopardy follow-up question: If the ball rolled off the platform twice as fast, would it stay in the air longer, shorter, or for the same amount of time?
The horizontal launch error
Correct the work
A student is calculating the fall time of a ball launched horizontally at 6.0 m/s from a height of 20 m.
They write:
-20=(6.0)t+1/2(-9.8)t^2
Identify the mistake and write the correct equation.
Answer: The student has incorrectly used the horizontal launch velocity as the initial vertical velocity.
Explanation: The ball begins with zero vertical velocity, even though it has horizontal velocity. The two components must be analyzed separately.
A student sketches a velocity–time graph for a ball thrown vertically upward and caught at its original height.
Their graph starts at a positive velocity, slopes downward until reaching zero at the highest point, and then becomes a horizontal line at zero.
The student claims:
"My graph is correct because the ball's velocity becomes zero at the top."
Are they right? Convince us by sketching the correct graph.
The correct velocity–time graph is a straight line with a constant negative slope:
Starts at a positive velocity (rising).
Crosses zero at the highest point.
Continues into negative velocity (falling).
Ends at the same speed magnitude as it started, but with negative velocity.
Why: Gravity continues accelerating the ball downward throughout its flight, including at the highest point.
Four graphs, one projectile
Question
A ball is launched upward at an angle and lands at its original height.
Sketch these four graphs for its entire flight:
Horizontal position vs. time
Vertical position vs. time
Horizontal velocity vs. time
Vertical velocity vs. time
Assume the launch point is the origin and upward/right are positive.
Answer
Straight line increasing from zero.
Downward-opening parabola, starting and ending at zero.
Horizontal line above zero.
Straight line decreasing from positive to negative, crossing zero at the top.
Each graph tells a different part of the same motion. The horizontal velocity remains constant while vertical velocity changes uniformly.
A red ball is fired straight into the air with an initial velocity of 10 m/s. One second later, an identical blue ball is fired in exactly the same way. The two balls will collide when the red ball
a. has reached its maximum altitude.
b. is going up and the blue ball is also going up.
c. is coming down and the blue ball is going up.
d. is coming down and the blue ball is coming down.
Answer: C
Gravity will have turned the red ball around just after a second and then the blue ball will be on its way up then.
Double the speed or double the height?
A ball rolls horizontally off a platform and lands some distance away.
Consider two changes:
A: Double its horizontal launch speed.
B: Double the platform height, keeping the launch speed the same.
Which change produces a greater horizontal range? Explain without using specific numbers.
Answer: Change A produces the greater range.
Explanation:
Horizontal range is:
Doubling horizontal speed doubles the range. Doubling height increases flight time and range by only sqrt(2) = 1.41
A cart moves right with constant positive acceleration. A student draws a straight, upward-sloping position–time graph and argues:
"The graph is increasing, so it shows the cart accelerating."
What is correct about the explanation, and what needs to change?
Answer: The increasing position correctly indicates motion in the positive direction. However, a straight position–time graph has constant slope and therefore constant velocity, not positive acceleration.
Explanation: To represent speeding up with constant positive acceleration, the position–time graph must curve upward, with its slope increasing over time.
A ball is launched from a cliff at an unknown angle.
You know its initial speed is 25 m/s and that it lands 80 m horizontally from the base of the cliff.
Can we determine the cliff's height?
A. Yes, using only horizontal motion.
B. Yes, using only vertical motion.
C. Yes, using the Pythagorean theorem.
D. No, we need more information.
Answer: D Not enough information for a unique height.
We don't know how the launch velocity is divided into horizontal and vertical components.
Knowing the launch angle, horizontal velocity, or flight time would provide enough additional information.
The police pursuit
Question:
A motorist passes a stationary police officer travelling at a constant 28 m/s.
The officer immediately accelerates from rest at 1.8 m/s².
How long does it take the officer to catch the motorist? Who is travelling faster when they meet?
Bonus Points if you draw their position vs time graphs.
Answer:
t = 31.1s and the officer is travelling faster
Key: Find when their displacements are equal + draw graph for visual help
2 objects can have the same position and 2 different speeds
A model rocket (bottle) launches vertically upward from rest.
While its engine is firing, it accelerates upward at a constant 12.0 m/s² for 5.0 seconds.
The engine then shuts off, and the rocket continues upward under the influence of gravity alone.
Ignoring air resistance, what maximum height does the rocket reach above its launch point?
Answer: 334 m
explanation: connecting variable is v when engine turns off find that first --> then find height at that point
then vertical projectile, vf = 0 --> find height gained after that point
height = total
A BMX rider leaves a cliff horizontally. We know the horizontal launch speed and the horizontal distance travelled before landing. We want to determine the cliff's height. Which direction would you investigate first, and why?
Answer: Start with the horizontal motion, because it provides the shared flight time. Then vertical motion gives the cliff height.
Key takeaway Start with the direction that contains enough information to determine something useful.
A ball launches at 20 m/s, at 30° above the horizontal, from a platform 15 m high.
A student tries to determine how far it travels horizontally.
They write:
vix=20cos30=17.3m/s
viy=20sin30=10.0m/s
At the top, vy=0
0=10.0-9.8t
t=1.02s
x=17.3(1.02)=17.7
The student concludes the ball lands 17.7 m from the platform.
What's wrong with the solution? Correct the approach.
The student correctly calculated the time to reach the highest point, but incorrectly treated that as the total flight time.
x = 52.8m
t = 3.05s
Two projectiles launched at the same speed, one at 30° and one at 60°, will travel exactly the same horizontal distance. Convince me that I'm right—or prove me wrong.
30 degrees has large vx but small vy so it has short flight time but covers distance
d = vx(big) x t(small)
60 degrees has small vx but large flight time
d = vx(small) x t(big)