Skip to contentDisplacement vs Distance
Displacement
- ฮx=xfโโxiโโ
- ฮx: Displacement
- xfโ: final position
- xiโ: initial position
- change in position of an object
- can be positive or negative
Distance
- the magnitude or size of displacement between two positions
- distance traveled is the total length of the path traveled between two positions
- distance is either more than ro equal to displacement
Scalar VS Vector
| Scalar | Vector |
|---|
| Magnitude | โ
| โ
|
| Direction | โ
| โ |
Time, Velocity, and Speed
Time
- ฮt=tfโโtiโ
- ฮt: change in time / elapsed time
- tfโ: time at the end of motion
- tiโ: time at the beginning of motion
Velocity
- vaveโ=ฮtฮxโ=tfโโtiโxfโโxiโโโ
- vaveโ: average velocity
- ฮx: displacement
- ฮt: elapsed time
- how fast an object travels in a given direction
- vector
Instantaneous Velocity
- v=ฮtโ0limโฮtฮxโ=dtdxโโ
- โvelocityโ usually means โinstantaneous velocityโ unless specified otherwise
Speed
- how fast/how much distance travelled in a given time interval
- Speed=timeย intervaldistanceย travelledย inย aย givenย timeย intervalโ
- scalar
Acceleration
- aaveโ=ฮtฮvโ=tfโโtiโvfโโviโโโ
- aaveโ: average acceleration
- ฮv: change in velocity
- ฮt: change in time
- vector
Kinematic Equations (Constant Acceleration)
- v=ฮtฮxโ=2vfโโviโโ
- vfโ=viโ+aฮt
- ฮx=xiโ+viโฮt+21โa(ฮt)2
- ฮx=21โ(viโ+vfโ)ฮt
- 2aฮx=vf2โโvi2โ
Graphs

Falling Objects
Gravity
- g=โ9.80m/s2โ (upward is position; downward is negative)
Kinematic Equations for Objects in Free-Fall:
- vfโ=viโ+gt
- yfโ=yiโ+viโฮt+21โg(ฮt)2
- 2aฮy=vf2โโvi2โ
Ball dropped from a tower
Question Suppose that a ball is dropped (v0โ=0) from atower. How far will it have fallen after a time:
- t1โ=1.00s
- t2โ=2.00s
- t3โ=3.00s
- Take y as positive downward
- so acceleration is a=g=+9.80m/s2
- v0โ=0m/s
- y0โ=0m
- Use Kinematic Equations for Objects in Free-Fall (2)
- yfโ=yiโ+viโฮt+21โg(ฮt)2
- t1โ=1.00s
- y1โ=0m+0m/s(1s)+21โ(9.8m/s2)(1s)2
- โดy1โ=4.9m
- t2โ=2.00s
- y1โ=0m+0m/s(2s)+21โ(9.8m/s2)(2s)2
- โดy1โ=19.6m
- t3โ=3.00s
- y1โ=0m+0m/s(3s)+21โ(9.8m/s2)(3s)2
- โดy1โ=44.1m
Ball thrown down from a tower
Question Suppose that a ball is thrown downward with an initial velocity of 3.00m/s
- What would be itโs position after 1s, 2s, 3s?
- What would its speed be after 1s, 2s?
Ball thrown upward
Question A person throws a ball upward into the air with an initial velocity of 15.0m/s. calculate how high it goes. Ignore air resistance.
- Take y as positive upward
- so a=โg=โ9.8m/s2
- As the ball rises, its speed decreases until it reaches the highest point, where its speed is zero for an instant; then it descends with increasing speed
- At t=0s:
- y0โ=0m
- v0โ=15m/s
- a=โ9.8m/s2
- At t=max
- vmโ=0
- a=โ9.8m/s2
- Use Kinematic Equation for Objects in Free-Fall (3)
- 2aฮy=vf2โโvi2โ
- 2gฮy=vm2โโv02โ
- 2(โ9.8m/s)(ymโโ0m)=(0m/s)2โ(15m/s)2
- ymโ=11.48959184m
- โ11.5m <3sf>
Ball thrown upward at edge of cliff
Vectors
Projectile Motion
โ ๏ธ EQUATIONS OF PROJECTILE MOTION
- vfโ=viโ+aฮt
- ฮx=viโฮt+21โa(ฮt)2
- ฮx=21โ(viโ+vfโ)ฮt
- 2aฮx=vf2โโvi2โ
Horizontal motion (x direction) [donโt need to remember]
- axโ=0
- (vxโ)fโ=(vxโ)iโ
- xfโ=xiโ+viโฮt
Vertical motion (y direction) [donโt need to remember]
- ayโ=โg
- (vyโ)fโ=(vyโ)iโโgฮt
- yfโ=yiโ+(vyโ)iโฮtโ21โg(ฮt)2
Textbook Problems
- Ch2: 3,6,9,10,15*,16*,18,20,25,31,35*,38,49*,57
- Ch3: 3,4,9,13,18,19,26,27,32,35*