Motion Problems

Constant V / Acceleration

If velocity is constant .... $a = $

Constant Acceleration Equations	Modify for Constant Velocity
$V_f = V_i + a\Delta t$	$V_f = $
$\Delta d = V_it + (\frac{1}{2})at^2$	$\Delta d = $
$V_f^2 = V_i^2 + 2a\Delta d$	$V_f = $
$\Delta d = \frac{[V_i + V_f]}{2}t$

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 ($s$).

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I. Speed and Velocity are Different

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 $\vec{v} = \Delta \vec{d} / \Delta t$.

Definitions: Displacement (direction, start); Speed (SCALAR); Velocity (VECTOR).

Logic: Velocity is a vector; a curve ball changes its velocity vector because its direction changes, even if its speed is a constant $35 \text{ m/s}$.

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II. Acceleration

Acceleration - time rate of change in

Which car has the greater acceleration?

T (sec)	t = 0	t = 1	t = 2	t = 3	t = 4	t = 5
Car A	70 m/s	72 m/s	74 m/s	76 m/s	78 m/s	80 m/s
Car B	0 m/s	3 m/s	6 m/s	9 m/s	12 m/s	15 m/s
Car C	90 m/s	90 m/s	90 m/s	90 m/s	90 m/s	90 m/s
Car D	100 m/s	99 m/s	98 m/s	97 m/s	96 m/s	95 m/s

a) Explain why each acceleration listed above could be called CONSTANT acceleration:

b) Under what condition is acceleration constant?

c) Show an example of an increasing acceleration given $V_i = 10 \text{ m/s}$:

10 m/s, 15 m/s, , ,

d) Show an example of a decreasing acceleration given $V_i = -100 \text{ m/s}$:

-100 m/s, -150 m/s, , ,

e) If an object’s acceleration is $-3 \text{ m/s}^2$, does that mean it is slowing down or speeding up? Give examples to prove your point:

f) If an object’s velocity is west and it is slowing down, the direction of the acceleration is Force is

g) There are 3 ways to have non-zero acceleration: , ,

Teacher Answer Key

Formula: $a = (V_f - V_i) / t$.

Calculations: Car A: $a = \mathbf{2 \text{ m/s}^2}$. Car B: $a = \mathbf{3 \text{ m/s}^2}$ (highest). Car C: $a = \mathbf{0 \text{ m/s}^2}$. Car D: $a = \mathbf{-1 \text{ m/s}^2}$.

Logic: Acceleration is constant if velocity changes by a uniform magnitude ($2 \text{ m/s}$, etc.) in every equal time interval.

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Acceleration Due to Gravity

Quick Review of Free Fall

Why the acceleration of a falling body does not depend on MASS: $a = F_{net}/m = $

a) A projectile with negligible friction has a acceleration of $m/s^2$ (direction)

b) When an object is falling, make down

c) When an object is ascending, make down

d) The velocity at the peak is

e) The vertical acceleration at the peak is and

f) When an object falls from rest, the relationship between its displacement and time is Equation that proves this

g) Compare the displacement after an object has fallen for 1 second with its displacement after 3 seconds

h) Compare the time it takes for an object to fall for 5 meters, with the time it takes to fall 10 meters

Ex 1) Tennis ball shot vertically upward from 165-meter cliff at $40 \text{ m/s}$. ($g=10 \text{ m/s}^2$)

How much time does it take for the ball to reach its peak?
How high above the base (bottom) of the cliff does the ball travel?
What is the ball’s vertical velocity 10 seconds after it is shot?
What is the ball’s vertical acceleration 2.5 seconds after it is shot?

Teacher Answer Key

Equations ($g = 10 \text{ m/s}^2$): 1. Peak time ($V_f = 0 \text{ m/s}$): $0 = 40 + (-10)t \implies \mathbf{4 \text{ s}}$.
2. Peak $\Delta d = V_it + 0.5at^2 = 40(4) - 5(16) = \mathbf{80 \text{ m}}$. Total: $165 \text{ m} + 80 \text{ m} = \mathbf{245 \text{ m}}$.
3. $V_f = 40 + (-10)(10) = \mathbf{-60 \text{ m/s}}$.
4. Acceleration is always constant at $\mathbf{-10 \text{ m/s}^2}$.

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 ($t^2$).

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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

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 ($y$): $d_y = 0.5gt^2$.

Solver ($g=10 \text{ m/s}^2$): 1. $t = \sqrt{80/10} = 2.83 \text{ s}$. $V_x = 45 \text{ m} / 2.83 \text{ s} = \mathbf{15.9 \text{ m/s}}$.
2. $t = 15 \text{ m} / 3.8 \text{ m/s} = 3.95 \text{ s}$. $d_y = 0.5(10)(3.95)^2 = \mathbf{78.0 \text{ m}}$.
5. $V_i = \sqrt{2 \cdot 10 \cdot 1940} = 197 \text{ m/s}$. $t = \sqrt{5/10} = 0.71 \text{ s}$. $R = 197 \cdot 0.71 = \mathbf{140 \text{ m}}$.

Logic: Perpendicular components are independent. Gravity only alters the vertical velocity ($m/s$), while horizontal velocity ($m/s$) remains constant.

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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.

V_ix =

V_iy =

Projectile – Level Ground Vx and Vy

$V_{ix} = 25 \text{ m/s} \quad V_{iy} = 40 \text{ m/s} \quad a = -10 \text{ m/s}^2$

a) Total time in the air = the peak time

b) V_iy at beginning =

What is the speed of the ball at 2 seconds?

Teacher Answer Key

Components @ 2s: $V_x = \mathbf{25 \text{ m/s}}$ (constant). $V_y = 40 + (-10)(2) = \mathbf{20 \text{ m/s}}$.

Resultant Speed: $V = \sqrt{V_x^2 + V_y^2} = \sqrt{(25)^2 + (20)^2} = \sqrt{625 + 400} = \mathbf{32.0 \text{ m/s}}$.

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 $= \mathbf{8 \text{ s}}$.

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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 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 ($m/s$).

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VII. Velocity vs Time Analysis

Describe these plots (image27, image13, image7):

1. A cart travels along a straight section of a road (image2):

• 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 (image23):

Describe motion and check return to start (image6, image18):

Teacher Answer Key

Area Rule: Displacement $\Delta d \text{ (m)} = \text{Area under } v-t \text{ curve}$.

Calculations (image23): Triangle Area (0-4s): $0.5(4 \text{ s})(10 \text{ m/s}) = \mathbf{20 \text{ m}}$. Rectangle Area (4-9s): $(5 \text{ s})(10 \text{ m/s}) = \mathbf{50 \text{ m}}$.

Final Position: $x_{final} = x_{initial} + \Delta d_{total} = 5 \text{ m} + 20 \text{ m} + 50 \text{ m} = \mathbf{75 \text{ m}}$.

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VIII. Position, Velocity, Acceleration – V vs t

Given the following plot that shows the velocity of an object as a function of time.

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

Area Formula: $\Delta d = \text{base} \cdot \text{height} \text{ (m)}$.

Intervals: 1. 0-2s Area $= 2 \text{ s} \cdot 5 \text{ m/s} = \mathbf{10 \text{ m}}$. Pos $= 5 + 10 = \mathbf{15 \text{ m}}$.
2. 2-6s Area $= 4 \text{ s} \cdot 10 \text{ m/s} = \mathbf{40 \text{ m}}$. Pos $= 15 + 40 = \mathbf{55 \text{ m}}$.

Logic: Area sums describe the cumulative position shift in meters. Acceleration is derived from the slope of these constant segments ($\mathbf{a = 0 \text{ m/s}^2}$).

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Show Me The PhysicsAP 1 Motion

Motion Problems

I. Speed and Velocity are Different

II. Acceleration

Which car has the greater acceleration?

Acceleration Due to Gravity

Object Propelled Horizontally

Practice Problems

Projectile Motion

VI. Intro to Motion Plots

VII. Velocity vs Time Analysis

VIII. Position, Velocity, Acceleration – V vs t

Show Me The Physics
AP 1 Motion