Position, velocity, and acceleration in one dimension
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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One-Dimensional Motion is part of the AP Physics 1 course on Study Mondo, specifically in the Kinematics section. You can explore the full course for more related topics and practice resources.
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Position and Displacement
Position
Position (x) is the location of an object relative to a reference point (origin).
Measured in meters (m)
Can be positive or negative
Example: x=5 m means 5 meters to the right of the origin
Displacement
Displacement (Δx) is the change in position.
Δx=xf−xi
Where:
xf = final position
xi = initial position
Note: Displacement is different from distance traveled!
Displacement: straight-line change (can be negative)
Distance: total path length (always positive)
Velocity
Average Velocity
Average velocity is displacement divided by time interval.
vavg=ΔtΔx=tf−tixf−x
Units: m/s
Can be positive (moving right/up) or negative (moving left/down)
Vector quantity (has direction)
Instantaneous Velocity
Instantaneous velocity is the velocity at a specific instant.
v=limΔt→0ΔtΔx=dtdx
This is the slope of the position-time graph at a point
On a graph: tangent line slope
Speed
Speed is the magnitude of velocity (always positive).
speed=∣v∣
Acceleration
Average Acceleration
Average acceleration is the change in velocity divided by time.
aavg=ΔtΔv=tf−tivf−v
Units: m/s²
Can be positive or negative
Positive a: speeding up in positive direction OR slowing down in negative direction
Instantaneous Acceleration
Instantaneous acceleration is the acceleration at a specific instant.
a=limΔt→0ΔtΔv=dtdv=dt2d2x
This is the slope of the velocity-time graph at a point
Kinematic Equations (Constant Acceleration)
When acceleration is constant, we can use these powerful equations:
The Big Four Kinematic Equations
v=v0+at
x=x0+v0t+21at2
v2=v02+2a(x−x0)
x=x0+21(v0+v)t
Where:
x0 = initial position
x = final position
v0 = initial velocity
v = final velocity
a = acceleration (constant)
t = time
Choosing the Right Equation
Missing variable
Use equation
x
v=v0+at
v
x=x0+v
a
x=x0+2
t
v2=v02
Free Fall
Free fall is motion under gravity alone (no air resistance).
a=−g=−9.8 m/s2
Negative because gravity pulls downward
Same equations apply, but use a=−g
Objects thrown upward: velocity decreases at rate g until v=0 at peak
Key Facts About Free Fall
At the peak of motion, v=0 but a=−g
Time up = time down (for same height)
Landing speed = launching speed (magnitude)
Sign Conventions
Always establish a coordinate system first!
Common convention:
Positive direction: right, up
Negative direction: left, down
Then:
Velocity in positive direction → v>0
Velocity in negative direction → v<0
Acceleration in positive direction → a>0
Acceleration in negative direction → a<0
Problem-Solving Strategy
Draw a diagram with coordinate system
List known variables (x0, v0, v, a, t)
Identify unknown you need to find
Choose equation that relates knowns and unknown
Solve algebraically before plugging in numbers
Check units and reasonableness
v0=0
Acceleration: a=2.0 m/s²
Time: t=5.0 s
Initial position: x0=0 (assume)
Find: Distance traveled (x)
Choose equation: We have v0, a, and t, and need x. Missing v.
Use: x=x0+v0t+21at2
Solve:x=0+(0)(5.0)+21(2.0)(5.0)2x=0+0+21(2.0)(25)x=21(50)x=25 m
Answer: The car travels 25 meters.
Check: Units are correct (m). A car accelerating for 5 seconds should travel a reasonable distance, and 25 m ≈ 82 feet seems right.
2Problem 2medium
❓ Question:
A car accelerates uniformly from rest to 25 m/s in 5.0 seconds. (a) What is the car's acceleration? (b) How far does the car travel during this time? (c) What is the car's average velocity during this interval?
💡 Show Solution
Solution:
Given: v₀ = 0 m/s (starts from rest), v = 25 m/s, t = 5.0 s
(a) Acceleration:
Using v = v₀ + at
25 = 0 + a(5.0)
a = 25/5.0 = 5.0 m/s²
(b) Distance traveled:
Using Δx = v₀t + ½at²
Δx = 0 + ½(5.0)(5.0)²
Δx = ½(5.0)(25)
Δx = 62.5 m or 63 m
Alternatively, using v² = v₀² + 2aΔx:
(25)² = 0 + 2(5.0)Δx
625 = 10Δx
Δx = 62.5 m ✓
(c) Average velocity:
v_avg = (v₀ + v)/2 = (0 + 25)/2 = 12.5 m/s
Or: v_avg = Δx/t = 62.5/5.0 = 12.5 m/s ✓
3Problem 3medium
❓ Question:
A ball is thrown vertically upward with an initial velocity of 20 m/s. How high does it go? (Use g=10 m/s²)
💡 Show Solution
Set up coordinate system:
Positive direction: upward
Origin: ground level
Given information:
Initial velocity: v0=+20 m/s (upward)
Acceleration: a=−g=−10 m/s² (gravity pulls down)
At maximum height: m/s
Find: Maximum height (y)
Choose equation: We have v0, v, and a, need y. Missing .
Use: v2=v02+2a(y−
Solve:02=(20)2+2(−10)(y−0)
Answer: The ball reaches a maximum height of 20 meters.
Physical insight: At the peak, velocity is zero but acceleration is still −10 m/s² (gravity never stops pulling down!).
4Problem 4hard
❓ Question:
A ball is thrown straight upward with an initial velocity of 20 m/s. Taking upward as positive and g = 10 m/s², find: (a) the maximum height reached, (b) the time to reach maximum height, (c) the total time in the air before returning to the starting point, and (d) the velocity when it returns to the starting height.
Method 1 - Use the time we just found:
x=x0+v0t+
Method 2 - Without using time (faster!):
v2=v02+2a(x−
Answers:
Time to stop: 20 seconds
Distance traveled: 300 meters
Check: Average velocity = 230+0=15 m/s. Distance = (15)(20)= m ✓
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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.