Transcript Physics 1A: Introduction to Physics and Problem Solving

Lecture 5:
Vectors &
Motion in 2 Dimensions
Questions of Yesterday
2) I drop ball A and it hits the ground at t1. I throw ball B
horizontally (v0y = 0) and it hits the ground at t2. Which is
correct?
a) t1 < t2
b) t1 > t2
c) t1 = t2
Questions of Yesterday
1) Can a vector A have a component greater than its magnitude
A?
a) YES
b) NO
2) What are the signs of the x- and y-components
of A + B in this figure?
a) (x,y) = (+,+)
b) (+,-)
c) (-,+)
d) (-,-)
Displacement in 2 Dimensions
Position vectors no longer accounted for by + and Displacement = change in position vector of object = Dr
Dx = xf - xi
Dr = rf - ri
1 Dimension
2 Dimensions
y (m)
ri
Dr
rf
object path
x (m)
Velocity in 2 Dimensions
Average Velocity
vav = Dr =
Dt
y (m)
rf - r i
tf - ti
ti
ri
Instantaneous Velocity
Dr
tf
rf
object path
x (m)
Dr
v = Dlim
t -> 0
Dt
Acceleration in 2 Dimensions
Average Acceleration
aav = Dv =
Dt
vy (m/s)
ti
vi
Instantaneous
Acceleration
Dv
tf
vf
vf - vi
tf - ti
object’s instant.
velocity
vx (m/s)
Dv
a = Dlim
t -> 0
Dt
Acceleration in 2 Dimensions
Average Acceleration
aav = Dv =
Dt
vf - vi
tf - ti
If a car is going North at a constant speed and makes a left
turn while maintaining its constant speed and then continues
West at the same speed…
does the car accelerate during this trip?
A runner is running on a circular track at constant speed?
Is she accelerating?
Acceleration in 2 Dimensions
Average Acceleration
aav = Dv =
Dt
vf - vi
tf - ti
Velocity is a vector with both magnitude & direction, so…
An object can accelerate by either
changing its SPEED or
changing its DIRECTION
Projectile Motion
Motion in 2 Dimensions under constant gravitational acceleration
Horizontal component of velocity is constant over entire path!
vx = v0x
No acceleration in horizontal direction
Projectile Motion
Motion in 2 Dimensions under constant gravitational acceleration
Vertical component of velocity constantly changing
due to gravitational acceleration in -y direction
v0y --> 0 -> -v0y
Projectile Motion
t= 1 s
t= 2 s
Initial velocity in horizontal direction
no gravity
t= 3 s
t= 4 s
Initial velocity in horizontal direction
with gravity
t= 5 s
Initial velocity in vertical direction with gravity
Horizontal and Vertical motions are
completely independent of each other!!
Motion in one direction has NO EFFECT
on motion in the other direction
Important Features of Projectile Motion
Acceleration is ALWAYS -9.80 m/s2 in the vertical direction
Parabolic motion is symmetric
vf = -v0
At the top of the trajectory:
t = 1/2 of total time
x = 1/2 of total horizontal range
t= 1 s
t= 2 s
t= 3 s
t= 4 s
t= 5 s
Total time of trajectory is
independent of horizontal motion
Important Features of Projectile Motion
At what point in the object’s trajectory is the
speed a minimum?
What about velocity?
t= 1 s
t= 2 s
t= 3 s
t= 4 s
t= 5 s
2D Motion under Constant Acceleration
Because x and y motions are independent…
we can apply 1D equations for constant acceleration motion
separately to each x- and y- direction
But….
v0 has both
x- and y-components
q
Need to separate v0 into
x- and y- components
v0x = v0cosq
v0y = v0sinq
2D Motion under Constant Acceleration
Recall equations for 1D motion under constant acceleration
v = v0 + at
Dx = v0t + 1/2at2
v2 = v02 + 2aDx
2D motion equivalent to superposition of two
independent motions in the x- and y-directions
Horizontal Motion
Vertical Motion
vx = v0x + axt
Dx = v0xt + 1/2axt2
vx2 = v0x2 + 2axDx
vy = v0y + ayt
Dy = v0yt + 1/2ayt2
vy2 = v0y2 + 2ayDy
Horizontal Motion of Projectile
vx = v0x + axt
Dx = v0xt + 1/2axt2
vx2 = v0x2 + 2axDx
vx = v0cosq = constant
Dx = v0xt = (v0cosq)t
v0x = v0cosq
ax = 0
t= 1 s
t= 2 s
t= 3 s
Time is still determined
by y-direction motion!
t= 4 s
t= 5 s
Vertical Motion of Projectile
vy = v0y + ayt
Dy = v0yt + 1/2ayt2
vy2 = v0y2 + 2ayDy
vy = v0sinq + gt
Dy = (v0sinq)t + 1/2gt2
vy2 = (v0sinq)2 + 2gDy
Pay attention to sign convention!
v0y = v0sinq
ay = g =
-9.80 m/s2
q
Equations for Motion of Projectile
Vertical
Component
Motion
vy = v0sinq + gt
Dy = (v0sinq)t + 1/2gt2
vy2 = (v0sinq)2 + 2gDy
Horizontal
Component
Motion
vx = v0cosq = constant
Dx = v0xt = (v0cosq)t
v = (vx2 + vy2)1/2
Combined
2D Motion
tanq = vy/vx
q = tan-1(vy/vx)
-90 < q < 90
Problem #1
A projectile falls beneath the straight-line path it would follow if
there were no gravity. How many meters does it fall below
this line if it has been traveling for 1 s? For 2 s?
Does your answer depend on the angle at which the projectile
is launched? What about the speed?
t= 1 s
t= 2 s
t= 3 s
t= 4 s
t= 5 s
Questions of the Day
2) Two projectiles are thrown with the same initial speed, one at
an angle q with respect to the ground and the other at an
angle 90o - q. Both projectiles strike the ground at the same
distance from the projection point. Are both projectiles in the
air for the same length of time?
a) YES
b) NO
1) A heavy crate is dropped from a high-flying airplane as it flies
directly over your shiny new car? Will your car get totaled?
a) YES
b) NO