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W1 Motion and Vectors

Displacement vs Distance

Displacement

  • ฮ”x=xfโˆ’xi\boxed{\Delta x = x_{f}-x_{i}}
    • ฮ”x\Delta x: Displacement
    • xfx_f: final position
    • xix_i: 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

ScalarVector
Magnitudeโœ…โœ…
Directionโœ…โŒ

Time, Velocity, and Speed

Time

  • ฮ”t=tfโˆ’ti\Delta t = t_f-t_i
  • ฮ”t\Delta t: change in time / elapsed time
  • tft_f: time at the end of motion
  • tit_i: time at the beginning of motion

Velocity

  • vave=ฮ”xฮ”t=xfโˆ’xitfโˆ’ti\boxed{v_{ave}=\frac{\Delta x}{\Delta t}=\frac{x_f-x_i}{t_f-t_i}}
    • vavev_{ave}: average velocity
    • ฮ”x\Delta x: displacement
    • ฮ”t\Delta t: elapsed time
  • how fast an object travels in a given direction
  • vector

Instantaneous Velocity

  • v=limโกฮ”tโ†’0ฮ”xฮ”t=dxdt\boxed{v=\lim_{\Delta t\rightarrow 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}}
  • โ€˜velocityโ€™ usually means โ€˜instantaneous velocityโ€™ unless specified otherwise

Speed

  • how fast/how much distance travelled in a given time interval
  • Speed=distanceย travelledย inย aย givenย timeย intervaltimeย interval\textmd{Speed}=\frac{\textmd{distance travelled in a given time interval}}{\textmd{time interval}}
  • scalar

Acceleration

  • aave=ฮ”vฮ”t=vfโˆ’vitfโˆ’ti\boxed{a_{ave}=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}}
    • aavea_{ave}: average acceleration
    • ฮ”v\Delta v: change in velocity
    • ฮ”t\Delta t: change in time
  • vector

Kinematic Equations (Constant Acceleration)

  1. v=ฮ”xฮ”t=vfโˆ’vi2v=\frac{\Delta x}{\Delta t}=\frac{v_f-v_i}{2}
  2. vf=vi+aฮ”tv_f = v_i + a\Delta t
  3. ฮ”x=xi+viฮ”t+12a(ฮ”t)2\Delta x = x_i + v_i\Delta t + \frac{1}{2}a(\Delta t)^2
  4. ฮ”x=12(vi+vf)ฮ”t\Delta x=\frac{1}{2}(v_i+v_f)\Delta t
  5. 2aฮ”x=vf2โˆ’vi22a\Delta x=v_f^2-v_i^2

Graphs

Falling Objects

Gravity

  • g=โˆ’9.80m/s2\boxed{g=-9.80 m/s^2} (upward is position; downward is negative)

Kinematic Equations for Objects in Free-Fall:

  1. vf=vi+gtv_f=v_i+gt
  2. yf=yi+viฮ”t+12g(ฮ”t)2y_f=y_i+v_i\Delta t+\frac{1}{2}g(\Delta t)^2
  3. 2aฮ”y=vf2โˆ’vi22a\Delta y=v_f^2-v_i^2

Ball dropped from a tower

  • Question Suppose that a ball is dropped (v0=0v_0=0) from atower. How far will it have fallen after a time:
    1. t1=1.00st_1=1.00s
    2. t2=2.00st_2=2.00s
    3. t3=3.00st_3=3.00s
  • Take yy as positive downward
  • so acceleration is a=g=+9.80m/s2a=g=+9.80m/s^2
  • v0=0m/sv_0=0m/s
  • y0=0my_0=0m
  • Use Kinematic Equations for Objects in Free-Fall (2)
    • yf=yi+viฮ”t+12g(ฮ”t)2y_f=y_i+v_i\Delta t+\frac{1}{2}g(\Delta t)^2
  1. t1=1.00st_1=1.00s
    • y1=0m+0m/s(1s)+12(9.8m/s2)(1s)2y_1=0m+0m/s(1s)+\frac{1}{2}(9.8m/s^2)(1s)^2
    • โˆดy1=4.9m\therefore y_1=4.9m
  2. t2=2.00st_2=2.00s
    • y1=0m+0m/s(2s)+12(9.8m/s2)(2s)2y_1=0m+0m/s(2s)+\frac{1}{2}(9.8m/s^2)(2s)^2
    • โˆดy1=19.6m\therefore y_1=19.6m
  3. t3=3.00st_3=3.00s
    • y1=0m+0m/s(3s)+12(9.8m/s2)(3s)2y_1=0m+0m/s(3s)+\frac{1}{2}(9.8m/s^2)(3s)^2
    • โˆดy1=44.1m\therefore y_1=44.1m

Ball thrown down from a tower

  • Question Suppose that a ball is thrown downward with an initial velocity of 3.00m/s3.00m/s
    1. What would be itโ€™s position after 1s, 2s, 3s?
    2. 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/s15.0m/s. calculate how high it goes. Ignore air resistance.
  • Take yy as positive upward
  • so a=โˆ’g=โˆ’9.8m/s2a=-g=-9.8m/s^2
  • 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=0st=0s:
    • y0=0my_0=0m
    • v0=15m/sv_0=15m/s
    • a=โˆ’9.8m/s2a=-9.8m/s^2
  • At t=maxt=max
    • vm=0v_m=0
    • a=โˆ’9.8m/s2a=-9.8m/s^2
  • Use Kinematic Equation for Objects in Free-Fall (3)
    • 2aฮ”y=vf2โˆ’vi22a\Delta y=v_f^2-v_i^2
  • 2gฮ”y=vm2โˆ’v022g\Delta y=v_m^2-v_0^2
  • 2(โˆ’9.8m/s)(ymโˆ’0m)=(0m/s)2โˆ’(15m/s)22(-9.8m/s)(y_m-0m)=(0m/s)^2-(15m/s)^2
  • ym=11.48959184my_m=11.48959184m
  • โ‰ˆ11.5m\approx 11.5m <3sf>

Ball thrown upward at edge of cliff

Vectors

Projectile Motion

โš ๏ธ EQUATIONS OF PROJECTILE MOTION

  • vf=vi+aฮ”tv_f=v_i+a\Delta t
  • ฮ”x=viฮ”t+12a(ฮ”t)2\Delta x=v_i\Delta t+\frac{1}{2}a(\Delta t)^2
  • ฮ”x=12(vi+vf)ฮ”t\Delta x=\frac{1}{2}(v_i+v_f)\Delta t
  • 2aฮ”x=vf2โˆ’vi22a\Delta x=v_f ^2-v_i ^2

Horizontal motion (xx direction) [donโ€™t need to remember]

  • ax=0a_x=0
  • (vx)f=(vx)i(v_x)_f=(v_x)_i
  • xf=xi+viฮ”tx_f=x_i+v_i\Delta t

Vertical motion (yy direction) [donโ€™t need to remember]

  • ay=โˆ’ga_y=-g
  • (vy)f=(vy)iโˆ’gฮ”t(v_y)_f=(v_y)_i-g\Delta t
  • yf=yi+(vy)iฮ”tโˆ’12g(ฮ”t)2y_f=y_i+(v_y)_i\Delta t-\frac{1}{2}g(\Delta t)^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*