CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
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95.141 Feb 6 , 2013 PHYSICS I Lecture 5
Register your i>clickers online at http://www.iclicker.com/registration
Last Lecture Chapter 3 Vector kinematics Projectile motion
Today Chapter 3 Beyond 1-D Vectors Relative Velocity
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
From two measurements of time and position
displacement
!x1, t1 !x2, t2!!x = !x2 "
!x1 a vector
Vector Kinematics: 1D
tip-to-tail
!x2 !!x1!
!x = !x2 "!x1
time interval !t = t2 " t1 a scalar
average velocity vavg =!!x!t
a vector
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Vector Kinematics •  Kinematics in more than one dimension •  Previously described 1D displacement as Δx, where
motion could only be positive or negative. •  In more than 1D, displacement is a vector
v!r
!r (t) !v(t)
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Vector Kinematics : 2D
12 xxx −=Δ
12 rrr −=Δ
x
y
!r1 !r2
!!r
displacement (in unit vector notation): !!r = (x2 " x1)i + (y2 " y1) j + (z2 " z1)k
average velocity = !!r!t
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Average to Instantaneous Just like in 1D, let Δt get smaller and smaller….
!v = lim!t" 0
!!r!t
=d!rdt
!!r !
!r
As Δt tends to zero, Δr tends to zero as well, but the ratio of Δr/Δt tends to a finite value Instantaneous velocity
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Instantaneous Velocity
Derivative of the position vector with respect to time:
!r = x(t)i + y(t) j + z(t)k
!v = d!rdt=dxdti + dy
dtj + dx
dtk
!v = vxi + vy j + vzk
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Acceleration Vector
Average acceleration =!!v!t
=!v2 "!v1
t2 " t1
!a = lim!t" 0
!!v!t
=d!vdt
Instantaneous acceleration
!a = d!vdt
=d 2xdt2
i + d2ydt2
j + d2zdt2
k=dvxdt
i +dvydt
j +dvydt
k
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Example Problem Given position as a function of time, find position, instantaneous velocity and instantaneous acceleration at t=3s
!v(t) = (8t)i + (2t)k
!a(t) = 8i + 2k
!r (t) = (4t2 !1)i + 2 j + (t2 )k
Plug in t=3s to get the final answers
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
By splitting the equations of motion into component form, we can solve problems one direction at a time An object starts from rest at the origin, with a constant acceleration in the +x direction of +2 m/s2 and in the –y direction of 1 m/s2.
What is the displacement of the object at t = 4 s ?
!x = v0 xt +12axt
2
!y = v0 yt +12ayt
2
Example Problem
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Motion with constant acceleration
Vector equations shorthand for separate scalar equations in x and y coordinates
tavvtatvrr +=++= 0
221
00
tavvtatvyy
tavvtatvxx
yyyyy
xxxxx
+=++=
+=++=
02
21
00
02
21
00
For projectile motion, ax = 0, ay = -g
gtvvgttvyyvvtvxx
yyy
xxx
−=−+=
=+=
02
21
00
000
Vector Kinematics
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Projectile Motion: Clicker Quiz
A helicopter moving at a constant horizontal velocity to the right drops a package when at position A. Which of the marked trajectories is closest to that observed by a stationary person on the ground?
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Projectile Motion
Red ball fired horizontally at
blue ball
Red ball fired at an angle towards
blue ball
Gravity OFF Gravity ON
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
•  Draw diagram, choose coordinate system
•  Knowns and unknowns •  Divide equations into x and y •  Solve, noting that in the x and
y calculations the common parameter is the time interval t
Helicopter flying horizontally at 70m/s wants to drop supplies on mountain top 200m below. How far in advance (horizontal distance) should the package be dropped?
Example (Rescue Helicopter)
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
t = 6.39s
x = 447m
x0 = 0 y0 = 0v0 x = 70 m/s v0 y = 0
ax = 0 ay = !9.8 m/s2
x = ? y = !200 m
y = yo + voyt +12ayt
2
common parameter is time t Solve y-equations to find t
Plug t into x-equations to find x
Example (Rescue Helicopter)
!200 = 0+ 0+ 12(!9.8)t2
x = xo + voxt +12axt
2
x = 0+ 70(6.39)+ 0
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Relative Motion: Clicker Quiz
A helicopter moving at a constant horizontal velocity to the right drops a package when at position A. Which of the marked trajectories is closest to that observed by a person on the helicopter?
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal. Find –  Equations of motion
•  Draw diagram and choose coordinate system •  Fill in knowns
x
y
v0
v0x
v0y θ0
x = xo + voxt +12axt
2
x0 = 0, y0 = 0v0 x = v0 cos!0, v0 y = v0 sin!0
ax = 0, ay = !g
Again, common parameter is time t
x = (vo cos!0 )t
y = yo + voyt +12ayt
2 y = (vo sin!0 )t !12gt2
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Example (Golf Ball)
A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal. –  The time of flight (how long the ball is in the air) This depends only on the y-component equations, as the acceleration due to gravity is only in the y-direction.
0 = voyt !12gt2y = yo + voyt +
12ayt
2 Since both y0 and y are zero
Two solutions for t where y=0
The second is the time of flight t =2v0 yg
0 = t(voy !gt2)
t = 0 and 2v0 y
g
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Example (Golf Ball)
R = 2vo2 sin!0 cos!o
g=vo2 sin2!0g
A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal. Find –  Range (how far does ball travel on flat ground)
Use constant x-velocity to calculate how far ball travels horizontally during time of flight (Range)
t =2v0 yg
=2v0 sin!
gtime of flight v0 x = v0 cos!0Constant
Range
Maximum Range occurs when sin(2θ0)=1 or θ0=45° Rmax =vo2
g
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Example (Golf Ball)
A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal. –  Trajectory (height y as a function of position x)
Since common parameter is time t, eliminate t to get y(x)
x = v0 xt y = voyt !12gt2
t = xv0 x
y =v0 yv0 x
!
"#
$
%&x '
g2v0 x
2
!
"#
$
%&x2
Equation of parabola y = Ax !Bx2
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
Example (Golf Ball)
The Speed Bus In the movie, a bus tries to jump a gap in the highway.
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
The Speed Bus
So we know: !vo,!x
1)  DRAW DIAGRAM!! 2)  Determine knowns 3)  Pick equations
t = !xv0 x
ov
y = ! 12gt2
= !12(9.8)*(0.5)2m
=15.24m31.3m / s
! 0.5s
!1.2m
Δx = 50m
Δy
= 31.3m/s
Bad news!
Is this even close to possible?
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
•  It looks like the bus is magically launched at an angle of 30º! •  Now can the bus make it?
But….
30º
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5
The Speed Bus
Speed Bus with Magic Launch
30º
ov
R = vo2 sin2!0g
Range
=(31.3)2 sin60°
9.81! 85m
CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5