EAS 215, Spring 2022
Assignment #1
SOLUTION
Part I: Practice with Vectors
1. Using the figure below,
a. Expressed in terms of i and j unit vectors, what is the vector result of 𝑣" + 𝑣$/" ?
b. Expressed in terms of i and j unit vectors, what is the vector result of 𝐹' + 𝐹( ?
c. What is the dot product of 𝑣" + 𝑣$/" ∙ 𝐹' + 𝐹( ?
d. What is the dot product of 𝐹( ∙ 𝑣$/" ?
!⃗#/% = 8
!⃗% = 10
!⃗%
60°
B
A
Part a:
𝑣" = 10𝑗
Part b:
&⃗' = 12
&⃗'
!⃗#/%
&⃗/ = 7
Y
X
&⃗/
𝑣$/" = 8 𝑖
𝒗𝑨 + 𝒗𝑩/𝑨 = 𝟖𝒊 + 𝟏𝟎𝒋
𝐹' = −12 cos 60° 𝑖 − 12 sin 60° 𝑗 = −6 𝑖 − 10.39 𝑗
𝐹( = −7 𝑗
𝑭𝟏 + 𝑭𝟐 = −𝟔 𝒊 − 𝟏𝟕. 𝟑𝟗 𝒋
Part c:
𝑣" + 𝑣$/" ∙ 𝐹' + 𝐹( = 8 −6 + (10)(−17.39)
𝒗𝑨 + 𝒗𝑩/𝑨 ∙ 𝑭𝟏 + 𝑭𝟐
= −𝟐𝟐𝟏. 𝟗
Part d:
𝐹( ⊥ 𝑣$/" So,
𝑭𝟐 ∙ 𝒗𝑩/𝑨 = 𝟎
2. Express the vector 𝑎 (below) in terms of i and j unit vectors.
Method 1
Find magnitude of 𝑎
Find total angle with x-axis, 𝜃
Use 𝑎 = 𝑎 cos 𝜃 𝑖 + 𝑎 sin 𝜃 𝑗
Y
3
2
$⃗
!
X
!"°
𝑎 =
2( + 3( = 3.606
2
𝜙 = atan( ) = 0.588 𝑟𝑎𝑑 = 33.69°
3
Total Angle 𝜃 = 𝜙 + 30° = 63.69°
𝑎 = 𝑎 cos 𝜃 𝑖 + 𝑎 sin 𝜃 𝑗
𝑎 = 3.606 cos 63.69° 𝑖 + 3.606 sin 63.69° 𝑗
𝒂 = 𝟏. 𝟔𝟎 𝒊 + 𝟑. 𝟐𝟑 𝒋
Method 2
Map components (3,2) into x,y space individually
Y
3
2
X
$⃗
!"°
=
3
!"°
+
2
12"°
𝑎 = 3 cos 30° 𝑖 + 3 sin 30° 𝑗 + 2 cos 120° 𝑖 + 2 sin 120° 𝑗
𝑎 = 2.598 𝑖 + 1.5 𝑗 + -1 𝑖 + 1.732 𝑗
𝒂 = 𝟏. 𝟔𝟎 𝒊 + 𝟑. 𝟐𝟑 𝒋
3. For each case below, what is the (scalar) component of vector 𝑎 in the direction of the
vector 𝑣 ?
a. 𝑎 = 45𝑖 + 44𝑗 + 10𝑘 and 𝑣 = 36𝑖 − 18𝑗 + 36𝑘
𝑎[ =
𝑎∙𝑣
45 36 + 44 −18 + (10)(36) 1188
=
=
𝑣
54
(36)( + (−18)( + (36)(
𝒂𝒗 = 22
Part b:
!⃗
120°
!⃗ = 15
#⃗ = 20
#⃗
Redrawn tail-to-tail, angle between vectors would be 𝜃 = 60°
𝑎[ = 𝑎 cos 60° = 15 cos 60°
𝒂𝒗 = 𝟕. 𝟓
Part c:
!⃗
120°
!⃗ = 20
#⃗ = 15
#⃗
Redrawn tail-to-tail, angle between vectors would be 𝜃 = 120°
𝑎[ = 𝑎 cos 120° = 20 cos 120°
𝒂𝒗 = −𝟏𝟎
4. In the figure below, what is the magnitude and direction of 𝑟 × 𝐹 ?
#⃗
30°
!⃗
!⃗ = 0.6
#⃗ = 200
Component of 𝐹 that is perpendicular to 𝑟 is 𝐹^ = 𝐹 cos 30°
𝑟 × 𝐹 = 𝑟 𝐹 cos 30° = 0.6 200 cos 30° = 103.9
By right hand rule, 𝑟 × 𝐹 is directed into the paper
𝒓 × 𝑭 = 𝟏𝟎𝟑. 𝟗 directed into the paper
OR, if converted to i,j,k:
𝒓 × 𝑭 = −𝟏𝟎𝟑. 𝟗 𝒌
5. For the figure below, what are the magnitudes and directions of the following cross
products?
a.
b.
c.
d.
𝑟' × 𝑚𝑎 = ?
𝑟( × 𝑚𝑎 = ?
𝑟' × 𝐹 = ?
𝑟( × 𝐹 = ?
B
!(
2
!"
2
A
m#⃗
&#⃗ = 160
%⃗ = 90
%⃗
3
3
Strategy:
Find components of 𝑟 that are perpendicular to the forces
Multiply by the magnitude of the force
Use right hand rule to get the direction
Part a:
𝑟' × 𝑚𝑎 = 2
160 = 𝟑𝟐𝟎 directed into the page
Part b:
𝑟( × 𝑚𝑎 = 2 160 = 𝟑𝟐𝟎 directed out of the page
Part c:
𝑟' × 𝐹 = 3 90 = 𝟐𝟕𝟎 directed into the page
Part d:
𝑟( 𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝐹 So, 𝒓𝟐 × 𝑭 = 𝟎
6. What is the vector cross product 𝑟 × 𝑣 ?
𝑟 = −2𝑖 + 4𝑗 − 4𝑘 and 𝑣 = 5𝑖 − 4𝑗 + 6𝑘
Use
𝑟×𝑣 = 𝑟h 𝑣i − 𝑟i 𝑣h 𝑖 + 𝑟i 𝑣j − 𝑟j 𝑣i 𝑗 + 𝑟j 𝑣h − 𝑟h 𝑣j 𝑘
Or remember,
i j k i j k i j k is positive
𝑟×𝑣 =
while reverse k j i k j i k j i is negative
4 6 − −4 −4 𝑖 + −4 5 − −2 6 𝑗 + −2 −4 − 4 5 𝑘
𝒓×𝒗 = 𝟖𝐢 − 𝟖𝐣 − 𝟏𝟐𝐤
Part II: Rectilinear (1-D) Motion
1. A ball is dropped from rest from a height of 3m above the ground.
a. What is the speed of the ball when it hits the ground?
b. How much time will pass between when it is released and when it hits the ground?
!" = 0
!
*"=0
s
% = 9.81 m/* +
3m
*+ = 3Define s=0 at release point, and s increasing as the ball drops.
Acceleration is then positive 𝑎 = 𝑔 = 9.81 m/𝑠 (
Part a
Constant acceleration so,
𝑣(( -𝑣o( = 2 𝑔 ∆𝑠
𝑣( =
𝑣o( + 2 𝑔 ∆𝑠 =
0 + 2 9.81 3
Part b
∆𝑣 = 𝑔 ∆𝑡 (constant acceleration)
𝒗𝟐 = 𝟕. 𝟔𝟕 𝒎/𝒔
∆𝑡 =
∆𝑣 7.67
=
𝑔
9.81
∆𝒕 = 𝟎. 𝟕𝟖 𝒔𝒆𝒄
2. A firetruck drives in a straight line with velocity shown on the graph below.
!
vel. [m/s]
!(#)
9
18
40
50
time [sec]
a. What is the acceleration of the firetruck during the three time intervals? 0s<time<18s?
18s<time<40s? 40s<time<50s?
b. What total distance did the firetruck travel during the entire 50 seconds?
Part a:
Interval 1: 0<t<18s: 𝒂 =
𝟗v𝟎
𝟏𝟖v𝟎
= 𝟎. 𝟓 m/𝒔𝟐
Interval 2: 18s<t<40s: 𝒂 = 𝟎 m/𝒔𝟐
𝟎v𝟗
Interval 3: 40s<t<50s: 𝒂 =
= −𝟎. 𝟗 m/𝒔𝟐
𝟓𝟎v𝟒𝟎
Part b:
!
vel. [m/s]
!(#)
9
1
2
18
3
40
50
time [sec]
xo
∆𝑠 =
o
𝑣 𝑑𝑡 = 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 = 𝐴' + 𝐴( + 𝐴~
'
'
∆𝑠 = ( 9 18-0 + 9 40 − 18 + ((9)(50 − 40)
∆𝑠 = 81 + 198 + 45
∆𝒔 = 𝟑𝟐𝟒 𝒎
3. A bee flies in a straight line (a “beeline” as they say). The initial speed of the bee is
𝑣o = 100 𝑐𝑚/𝑠. The acceleration of the bee versus position is given by the graph
below.
!
! [cm/" # ]
500
145
0
50
105
195
s [cm]
-250
a. At what position will the speed of the bee be maximum? Hint: it will be one of the
labeled points: 0cm, 50cm, 105cm, 145cm or 195cm.
b. What is the speed of the bee at that location?
c. What is the speed of the bee after traveling the full 195cm? Hints: you don’t need to
figure the equations for the four lines to do the integrals, and be careful of the signs.
Part a:
speed will increase as long as acceleration is positive
speed will decrease as long as acceleration is negative
Max speed will occur when acceleration crosses zero. So,
𝒗𝒎𝒂𝒙 𝒂𝒕 𝒔 = 𝟏𝟎𝟓𝒄𝒎 (or 1.05m)
Part b:
!
! [cm/" # ]
500
0
145
50
105
s [cm]
195
-250
(
𝑣'ox
− 𝑣o( = 2
'ox
𝑎 𝑑𝑠 = 2 ∗ (𝐴𝑟𝑒𝑎 𝑢𝑛𝑑 𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒)
o
𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 = •‚ 500 105 = 26250
𝑣'ox =
𝑣o( + 2 (𝐴𝑟𝑒𝑎) =
(100)( + 2 (26250)
𝒗𝟏𝟎𝟓 = 𝟐𝟓𝟎 𝒄𝒎/𝒔 (or 2.50 m/s)
Part c:
!
! [cm/" # ]
500
145
0
50
105
195
s [cm]
-250
(
𝑣'ƒx
−
(
𝑣'ox
'ƒx
=2
𝑎 𝑑𝑠 = 2 ∗ (𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒)
'ox
𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 = •‚ −250 195 − 105 = −11250
𝑣'ox =
(
𝑣'ox
+ 2 (𝐴𝑟𝑒𝑎) =
(250)( + 2 (−11250)
𝒗𝟏𝟎𝟓 = 𝟐𝟎𝟎 𝒄𝒎/𝒔 (or 2.00 m/s)