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 ($t^2$).
Object Propelled Horizontally
1. Whenever an object is projected horizontally:
a) $V_{iy} = $ b) $a_y = $ c) $a_x = $
d) $V_y$ + or – (circle) e) $\Delta d_y$ + or – (circle) f) $V_x$ is
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Find the speed at 3 seconds:
2. When solving problems in 2 dimensions, we must the x info from the
Practice Problems
1. Bob throws ball horizontally off 40.0-meter cliff. Hits ground 45.0 meters from base. How fast?
2. Erica throws ball at 3.80 m/s. If ball hits ground 15.0 m away, how high is the window?
3. Debbie shoots pellet horizontally at 12.5 m/s from top of 30.0-meter building. How far?
4. Louise drives car off 60.0 m cliff. Hits ground 125 m out. How fast?
5. Julia shoots BB gun straight up; reaches height of 1940 m. If shot horizontally from 2.50 m, how far?
Teacher Answer Key
Formula ($x$): $d_x = V_x \cdot t$. Formula ($y$): $d_y = 0.5gt^2$.
Logic: Perpendicular components are independent. Gravity only alters the vertical velocity ($m/s$), while horizontal velocity ($m/s$) remains constant.
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.
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 $= \mathbf{8 \text{ s}}$.
VI. Intro to Motion Plots
The key to a deep understanding of motion plots is to look at a plot's SLOPE.
SLOPE of a d vs t plot =
SLOPE of a V vs t plot =
Ex 1) Describe velocity, displacement, and acceleration of these plots:
Ex 2) Compare the 2 velocities shown in this
D vs t plot:
Ex 3) When is the object accelerating? Does the car ever move backward?
Ex 4) When velocity is changing the to the curve gives you the INSTANTANEOUS Velocity.
Velocity at 6 seconds / Slope of tangent =
Teacher Answer Key
Rule: Slope of position graph = Velocity (m/s). Slope of velocity graph = Acceleration (m/s²).
Ex 3: Line is curved, so slope is changing. Changing velocity = Acceleration. Since $d$ only increases, it never moves backward.
Logic: Backward motion requires a negative slope (decreasing $d$ in meters). Since the graph always trends up, velocity is always positive ($m/s$).
VII. Velocity vs Time Analysis
Describe these plots:
1. A cart travels along a straight section of a road:
• Indicate every time $t$ for which cart is at rest:
• Indicate interval where speed increases:
• Accel a-b: Accel b-d:
2. A 0.50 kg cart moves on a straight horizontal track:
Describe motion and check return to start:
Teacher Answer Key
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.
Area Rule: Displacement $\Delta d \text{ (m)} = \text{Area under } v-t \text{ curve}$.
Given the following plot that shows the velocity of an object as a function of time.
Object has d = 5.0 m at the start
Describe motion ($x_i = +5 \text{ m}$):
Determine $\Delta d$ for first two seconds:
Determine $\Delta d$ for next four seconds:
Determine $\Delta d$ for last two seconds:
Determine position at end of each interval:
Plot d vs. t. and a vs. t. on the grid below.
Teacher Answer Key
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.
