Concept 1) A car on cruise control moves at a constant 25 m/s. What is its acceleration? Which equation would you use to find how far it travels in 8 s?
Start with $\Delta d = V_i t + \tfrac{1}{2}at^2$. With $a = 0$ it becomes $\Delta d = vt$
Substitute: $\Delta d = (25 \text{ m/s})(8 \text{ s})$
$\Delta d = 200 \text{ m}$
Concept 2) If an object's acceleration is zero, can it still be moving? If an object's velocity is zero at one instant, must its acceleration be zero? (Think: a ball at the top of its path.)
Concept 3) Car A passes a stop sign at a constant 20 m/s. At that same instant Car B starts from rest at the sign and accelerates at 4 m/s². When do the cars have the same velocity? Has Car B caught up to Car A at that moment? Explain.
Given: Car A $v_A = 20 \text{ m/s}$ (constant). Car B $V_i = 0$, $a = 4 \text{ m/s}^2$
Same velocity when $V_B = 20 \text{ m/s}$: $V_f = V_i + at \;\Rightarrow\; 20 = 0 + (4)t$
$t = \dfrac{20}{4} = 5 \text{ s}$
Car A at 5 s: $\Delta d_A = vt = (20 \text{ m/s})(5 \text{ s}) = 100 \text{ m}$
Car B at 5 s: $\Delta d_B = V_i t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(4)(5)^2 = 50 \text{ m}$
No: they have the same velocity at 5 s, but A is 50 m ahead, so B has not caught up yet
Concept 4) For constant velocity, $V_f^2 = V_i^2 + 2a\Delta d$ reduces to $V_f = V_i$. Why is this equation no help for finding distance when $a = 0$?
Teacher Answer Key
Analysis: If velocity is constant, $\mathbf{a = 0 \text{ m/s}^2}$.
Equations: Using $a = 0 \text{ m/s}^2$ in kinematic formulas:
1. $V_f = V_i + (0)t \implies \mathbf{V_f = V_i}$.
2. $\Delta d = V_it + 0.5(0)t^2 \implies \mathbf{\Delta d = V \cdot t}$.
Logic: Constant velocity means zero acceleration, reducing the displacement formula to a direct product of velocity ($m/s$) and time ($s$).
Concept 1: $a = 0 \text{ m/s}^2$. Use $\Delta d = v t$ $= (25 \text{ m/s})(8 \text{ s})$ $= 200 \text{ m}$.
Concept 2: YES – zero acceleration means constant velocity, not zero velocity. NO – a ball thrown up has $v = 0 \text{ m/s}$ at the top, but $a = 9.8 \text{ m/s}^2$ down the whole time.
Concept 3: Same velocity when $V_f = at$: $20 \text{ m/s} = (4 \text{ m/s}^2)t$ $\Rightarrow t = 5 \text{ s}$. NO – at 5 s, A has gone $(20 \text{ m/s})(5 \text{ s}) = 100 \text{ m}$ and B only $\frac{1}{2}(4 \text{ m/s}^2)(5 \text{ s})^2 = 50 \text{ m}$. B was slower the whole time until then, so A is still ahead. (B catches A at $t = 10 \text{ s}$, 200 m from the sign.)
Concept 4: With $a = 0$ the $2a\Delta d$ term is zero, so $\Delta d$ drops out of the equation – it no longer contains distance. Use $\Delta d = v t$ instead.
Quick Check
3 questions on Equations — pick an answer to check yourself.
1. An object moves with constant velocity. What is its acceleration?
2. A car cruises at a constant 15 m/s for 4.0 s. How far does it travel?
3. A runner moves at constant velocity. Which statement is true?
I. Speed and Velocity are Different
Animation: Velocity vs Speed (Curve Ball)
a) Displacement – Distance and from to finish
b) Speed = Distance/time VECTOR or SCALAR???
c) Velocity is speed in a VECTOR or SCALAR
Ex) A curve ball has one speed but velocities because it moves
d) Velocity Equation =
Teacher Answer Key
Equation: Average Velocity $\vec{v} = \Delta \vec{d} / \Delta t$.
Ratio (g): $\Delta d = 0.5gt^2$. At $3 \text{ s}$, distance is $0.5(10)(9) = 45 \text{ m}$. At $1 \text{ s}$, distance is $5 \text{ m}$. It falls 9 times further.
Logic: Gravity affects all objects equally regardless of mass; the ratio of displacement is proportional to the square of the time ($t^2$).
NYSSLS Reading Passage · 1D Motion
The Physics of the Plunge: Force, Acceleration, and Terminal Velocity
Ex 2) A sphere is dropped from the same height as 3 spheres are projected horizontally. Show where the 3 spheres will be each second.
2. When solving problems in 2 dimensions, we must the x info from the
Object Projected Horizontally Equations
Equation (modify for X and Y)
X Direction $a_x = $ $V_x$ is
Y Direction $a_y = $ $V_{iy} = $
$\Delta d = \dfrac{[V_f + V_i]}{2}t$
$\Delta d = V_it + \tfrac{1}{2}at^2$
$V_f = V_i + at$
$V_f^2 = V_i^2 + 2a\Delta d$
Practice Problems
1. Bob throws a lacrosse ball horizontally off a 40.0-meter-high cliff. How fast did he throw the lacrosse ball if it hit the ground 45.0 meters from the base of the cliff?
Logic: Perpendicular components are independent. Gravity only alters the vertical velocity ($m/s$), while horizontal velocity ($m/s$) remains constant.
Quick Check
3 questions on Horizontal Projectiles — pick an answer to check yourself.
1. A ball is thrown horizontally at 30 m/s off a cliff. Ignoring air resistance, what is its horizontal velocity 3.0 s later?
2. A ball rolls horizontally off a 20 m high cliff at 5.0 m/s. How far from the base does it land? (g = 10 m/s²)
3. From the same height, one ball is dropped while another is thrown horizontally very fast. Ignoring air resistance, which lands first?
Projectile Motion
Range - Show it on the drawing above.
Solving Projectile using Components: We always solve projectile problems by resolving the velocity vectors into x and y components. Draw and label Vx and Vy in the picture below.
B) $100 \text{ m/s}$ at $30^\circ$ below the horizontal (same speed and angle as launch, on level ground)
Ex 2) $V_{iy} = 0$, $\;V_x = 26.25 \text{ m/s}$, $\;d_y = 61.25 \text{ m}$. Find the time in the air, the range, and the magnitude and direction of the velocity when the ball hits the ground.
$43.8 \text{ m/s}$ at $53.1^\circ$ below the horizontal
Ex 3) A ball is launched at $50 \text{ m/s}$ at $53.13^\circ$ above the horizontal from the top of a $165 \text{ m}$ cliff. Find the time in the air, the range, and the magnitude and direction of the velocity when the ball hits the ground.
Logic: Launch $V_{iy} = 40 \text{ m/s}$ takes $\mathbf{4 \text{ s}}$ to reach peak where $V_y = 0 \text{ m/s}$. Total time is symmetric, so total $= \mathbf{8 \text{ s}}$.
Level Ground Ex 1: $V_x = 86.6 \text{ m/s}$, $V_{iy} = 50.0 \text{ m/s}$. $t_{peak} = 5.00 \text{ s}$, total $= 10.0 \text{ s}$. Range $= (86.6)(10.0) = \mathbf{866 \text{ m}}$. Lands at $\mathbf{100 \text{ m/s}, \ 30^\circ}$ below the horizontal.
Ex 2: $t = \sqrt{2(61.25)/10} = \mathbf{3.50 \text{ s}}$. Range $= (26.25)(3.50) = \mathbf{91.9 \text{ m}}$. $V_{fy} = 35.0 \text{ m/s}$, so $v = \mathbf{43.8 \text{ m/s}}$ at $\mathbf{53.1^\circ}$ below the horizontal.
Ex 3: $V_x = 30.0 \text{ m/s}$, $V_{iy} = 40.0 \text{ m/s}$. $-165 = 40t - 5t^2 \Rightarrow t = \mathbf{11.0 \text{ s}}$. Range $= \mathbf{330 \text{ m}}$. $V_{fy} = -70.0 \text{ m/s}$, so $v = \mathbf{76.2 \text{ m/s}}$ at $\mathbf{66.8^\circ}$ below the horizontal.
Quick Check
3 questions on Angled Projectiles — pick an answer to check yourself.
1. At the very top of a projectile's path over level ground (no air resistance), which statement is true?
2. A ball is launched over level ground with Vix = 25 m/s and Viy = 40 m/s. Using a = −10 m/s², what is its total time in the air?
3. Same ball (Vix = 25 m/s, Viy = 40 m/s, a = −10 m/s²). What is its speed at t = 2.0 s?
Ex 5) Describe the motion that produced this plot.
Ex 6)
a) Rank the accelerations from largest to smallest.
b) Explain D
Teacher Answer Key
Rule: Slope of position graph = Velocity (m/s). Slope of velocity graph = Acceleration (m/s²).
Ex 3: From A to B the line is straight, so the slope (velocity) is constant and $a = 0$. After B the curve bends, so the slope is changing. Changing velocity = Acceleration: around C, D and E, and again after $t = 4 \text{ s}$ as the curve gets steeper. Between C and D, $d$ decreases, so the car does move backward briefly.
Logic: Backward motion requires a negative slope (decreasing $d$). The graph slopes down only between C and D, so that is the only time the velocity is negative and the car moves backward. Everywhere else the slope is positive or zero.
Quick Check
3 questions on Motion Plots — pick an answer to check yourself.
1. What does the slope of a velocity vs. time graph represent?
2. A straight-line position vs. time graph goes from 2 m at t = 0 to 14 m at t = 4.0 s. What is the object's velocity?
3. A position vs. time graph always rises and curves upward, getting steeper over time. What is the object doing?
2. A 0.50 kg cart moves on a straight horizontal track:
(a) Indicate every time $t$ for which the cart is at rest:
(b) Indicate every time interval for which the cart's speed (magnitude of velocity) is increasing:
(c) Determine the horizontal position $x$ of the cart at $t = 9.0$ s if the cart is located at $x = 0$ at $t = 0$:
Displacement = area under the $v$-$t$ graph (area below the axis is negative)
Return to Start: After $4 \text{ s}$ the object moves at a constant $+10 \text{ m/s}$, so it covers the $10 \text{ m}$ back to the start in $1 \text{ s}$: it returns to the starting point at $\mathbf{t = 5 \text{ s}}$. (Area $4\text{-}5 \text{ s}$: $(1 \text{ s})(10 \text{ m/s}) = +10 \text{ m}$, so $\Delta d_{total} = 0$.)
Claim (Graph 1 - Cart): The cart is at rest at $\mathbf{t = 0 \text{ s}, \ 5.0 \text{ s}, \text{ and } 13 \text{ s}}$ (points a, c, e). Speed increases over $\mathbf{0\text{-}3.0 \text{ s}, \ 5.0\text{-}8.0 \text{ s}, \text{ and } 13\text{-}18 \text{ s}}$. $a_{a\text{-}b} = \mathbf{1.0 \text{ m/s}^2}$ and $a_{b\text{-}d} = \mathbf{-1.6 \text{ m/s}^2}$.
Evidence (Graph 1 - Cart): $V = 0 \text{ m/s}$ only where the line crosses the T-axis: $t = 0, 5.0, 13 \text{ s}$. $|V|$ grows away from zero on three legs: a→b ($0 \to 3.0 \text{ m/s}$), c→d ($0 \to -5.0 \text{ m/s}$), and e→f ($0 \to 5.0 \text{ m/s}$). $a_{a\text{-}b} = \frac{3.0 - 0}{3.0 - 0} = \mathbf{1.0 \text{ m/s}^2}$. $a_{b\text{-}d} = \frac{-5.0 - 3.0}{8.0 - 3.0} = \frac{-8.0}{5.0} = \mathbf{-1.6 \text{ m/s}^2}$.
Reasoning (Graph 1 - Cart): "At rest" means $V = 0 \text{ m/s}$, which happens only at the T-axis crossings. Speed (the magnitude of $V$) increases whenever the line moves away from the T-axis, whether $V$ is positive or negative, so a-b, c-d, and e-f all count even though b-d passes through zero at c. Acceleration is the constant slope $\Delta V / \Delta t$ of a straight segment, so the single value $-1.6 \text{ m/s}^2$ describes the whole b-d segment despite the direction reversal at c.
Claim (Graph 2 - 0.50 kg Cart): (a) The cart is at rest at $\mathbf{t = 4.0 \text{ s and } 18 \text{ s}}$. (b) Speed increases over $\mathbf{4.0\text{-}9.0 \text{ s}}$ and $\mathbf{18\text{-}20 \text{ s}}$. (c) From $t = 0$ to $9.0 \text{ s}$ the displacement is $\mathbf{\Delta x = -0.90 \text{ m}}$, and the cart starts at $x = 0$, so at $t = 9.0 \text{ s}$ it is at $\mathbf{x = -0.90 \text{ m}}$ (0.90 m behind its starting point).
Evidence (Graph 2 - 0.50 kg Cart): $v_x = 0$ where the line crosses the $t$-axis: at $4.0 \text{ s}$ (falling from $+0.8$ to $-1.0 \text{ m/s}$ with slope $-0.20 \text{ m/s}^2$, so $0.8/0.20 = 4.0 \text{ s}$) and at $18 \text{ s}$ (rising from $-0.4$ to $+0.8 \text{ m/s}$ with slope $+0.40 \text{ m/s}^2$, so $17 + 0.4/0.40 = 18 \text{ s}$). $|v_x|$ grows from $0$ to $1.0 \text{ m/s}$ over $4.0\text{-}9.0 \text{ s}$ and from $0$ to $0.8 \text{ m/s}$ over $18\text{-}20 \text{ s}$. Area $0\text{-}4.0 \text{ s}$: $\tfrac{1}{2}(4.0)(0.8) = +1.6 \text{ m}$. Area $4.0\text{-}9.0 \text{ s}$: $\tfrac{1}{2}(5.0)(-1.0) = -2.5 \text{ m}$. Total $= \mathbf{-0.90 \text{ m}}$.
Reasoning (Graph 2 - 0.50 kg Cart): At rest means $v_x = 0$, which happens only at the axis crossings. Speed increases when the line moves away from the $t$-axis, whether $v_x$ is positive or negative; from $12\text{-}17 \text{ s}$ and $20\text{-}25 \text{ s}$ the line is flat, so speed is constant there. Displacement is the area under the $v$-$t$ graph, with area below the axis counting as negative, so the cart moves forward $1.6 \text{ m}$, then back $2.5 \text{ m}$, ending $0.90 \text{ m}$ behind where it started: $x = 0 + (-0.90 \text{ m}) = -0.90 \text{ m}$.
Quick Check
3 questions on Graph Analysis — pick an answer to check yourself.
1. On a velocity–time graph, when is an object's speed increasing?
2. A cart's velocity changes steadily from +3.0 m/s at t = 3.0 s to −5.0 m/s at t = 8.0 s. What is its acceleration?
3. A cart speeds up uniformly from rest to 10 m/s in 4.0 s, then moves at a constant 10 m/s for 5.0 s. What is its total displacement?
Claim: Starting at $x_i = +5 \text{ m}$ with $V_i = -10 \text{ m/s}$, the object has constant acceleration $\mathbf{a = 5 \text{ m/s}^2}$ from $t=0$ to $t=6 \text{ s}$ (it momentarily stops and reverses direction at $t = 2 \text{ s}$), then moves at constant velocity ($\mathbf{a = 0 \text{ m/s}^2}$) from $t=6 \text{ s}$ to $t=8 \text{ s}$, ending at $x = \mathbf{75 \text{ m}}$.
Area Formula: $\Delta d = \text{Area under } v\text{-}t \text{ curve}$. The line is one continuous slope from $(0,-10)$ to $(6,20)$, so $a = \Delta V/\Delta t = 30/6 = \mathbf{5 \text{ m/s}^2}$ the whole way, even though $V$ crosses $0$ at $t=2 \text{ s}$.
1. $0\text{-}2 \text{ s}$ (triangle below axis): $\Delta d = 0.5(2 \text{ s})(-10 \text{ m/s}) = \mathbf{-10 \text{ m}}$. Pos $= 5 - 10 = \mathbf{-5 \text{ m}}$.
2. $2\text{-}6 \text{ s}$ (triangle above axis): $\Delta d = 0.5(4 \text{ s})(20 \text{ m/s}) = \mathbf{40 \text{ m}}$. Pos $= -5 + 40 = \mathbf{35 \text{ m}}$.
3. $6\text{-}8 \text{ s}$ (rectangle, $a=0$): $\Delta d = (2 \text{ s})(20 \text{ m/s}) = \mathbf{40 \text{ m}}$. Pos $= 35 + 40 = \mathbf{75 \text{ m}}$.
Logic: A single straight slope means constant acceleration even as $V$ passes through zero — that crossing is just a turnaround (direction change), not a change in $a$. Area below the $t$-axis is negative (backward) displacement, which is why the object ends up at $x=-5 \text{ m}$ (behind its $+5 \text{ m}$ start) before the $2\text{-}6 \text{ s}$ and $6\text{-}8 \text{ s}$ intervals push it forward to $\mathbf{75 \text{ m}}$. From $t=6$ to $8 \text{ s}$ the graph is flat, so $a=0 \text{ m/s}^2$ there.
NYSSLS Reading Passage · 2D Motion
Newton's Cannonball: The Thought Experiment That Explained Orbits
The motion of a particle along a straight line is represented by the position versus time graph above. At which of the labeled points on the graph is the magnitude of the acceleration of the particle greatest?
Check Your Understanding
5 quick questions — pick an answer to see if you’re right.
1. An object moves at a constant 15 m/s for 4.0 s. Using the kinematic equations with a = 0, what is its displacement?
2. A ball is shot straight up at 20 m/s from the top of a 100 m cliff (g = 10 m/s²). How long does it take to reach its peak?
3. For the same ball (20 m/s up from a 100 m cliff, g = 10 m/s²), how high above the base of the cliff is its peak?
4. A ball is launched horizontally at 30 m/s (g = 10 m/s²). What is its speed 3.0 s later?
5. An object's acceleration is −3 m/s². Is it slowing down?
