Unlocking the Potential of the Cloud for IBM Power Systems
Kinematics equations worksheet
1. ANSWERS
1. d=? vf2 = vi2 + 2ad => d = vf2 - vi2
vi = 8 2a
vf = 10 => d = 102 - 82
a=6 2x6
t = =3m
2 (a) d = 400 vf2 = vi2 + 2ad => a = vf2 - vi2
vi = 25 2d
vf = 15 => a = 152 - 252
a=? 2 x 400
t =
= -0.5 ms-2
(b) d = (vi + vf).t d = 2d d = 2 x 400
=> =>
2 (vi + vf) (25 + 15)
= 20 s
Monday, 15 March 2010
2. 3 (a) d= t = vf - vi t = 18 - 12
vf = vi + at => =>
vi = 12 a 1.5
vf = 18 =4s
a = 1.5
t =?
(b) d=? d = vf2 - vi2 => d = 182 - 122
vi = 12 2a 2 x 1.5
vf = 18
a = 1.5 d = 60 m
t =4
(c) d = 100 vf2 = vi2 + 2ad => vf2 = 122 + 2 x 1.5 x 100
vi = 12 = 444
vf = ? vf =√ 444
a = 1.5
vf = 21.1 ms-1
t =
Monday, 15 March 2010
3. 4 (a) During this reaction time the vehicle travels with constant speed (a = 0) so:
v=d => d = v.t = 12 x 0.75 = 9 m
t
(b) d=? d = vf2 - vi2
vf2 2
= vi + 2ad =>
vi = 12 2a
vf = 0 d = 02 - 122
a = -2 2 x -2
t =
d = 36 m
(c) The dog survives
5. Consider the downward direction as positive and that the package will have the
same vertical acceleration as if it were dropped from a cliff at the same altitude.
(a) d = 1200
vf2 = vi2 + 2ad => vf2 = 02 + 2 X 10 X 1200
vi = 0 = 24000
vf = ?
a = 10 => vf =√ 24000
t = = 155 ms-1
Monday, 15 March 2010
4. 5 (b) t=? d = vit + 1at2 but vit = 0
2
t = 2d = 2 x 1200
= 15.5 s
a 10
6 (a) d = 120 d = vit + 1at2 but vit = 0
vi = 0 2
vf =
+ t = 2d = 2 x 120
a = 10 = 4.9 s
a 10
t =?
(b) vf = vi + at = 0 + 10 x 4.9 = 49 ms-1
(c) t = vf - vi t = 35 - 0
d= vf = vi + at => =>
a 10
vi = 0
= 3.5 s
vf = 35
a = 10
t =?
Monday, 15 March 2010
5. 7 (a) The flight path is as follows:
0 ms-1 For the 2nd half of the flight:
+
d= vf = vi + at
- vi = 0
vf = -96 => t = vf - vi
a = -10 a
96 ms-1 96 ms-1 t =? => t = 0 - 96
-10
There is the 1st half (travelling
up) and the 2nd half (travelling = 9.6 s
down). THE FLIGHT PATH IS
SYMMETRICAL
(b)
We are now d= vf = vi + at
analysing the vi = 96
1st half of the vf = 48 => t = vf - vi
flight a = -10 a
t =? => t = 48 - 96
-10
= 4.8 s
Monday, 15 March 2010
6. (c) We are still
d= vf = vi + at
analysing the vi = 96
1st half of the vf = 0 => t = vf - vi
flight a = -10 a
t =? => t = 0 - 96
-10
= 9.6 s
(d) We are still d=? d = vf2 - vi2
vf2 2
= vi + 2ad =>
analysing the vi = 96 2a
1st half of the vf = 0
d = 02 - 962
flight a = -10 2 x -10
t =
d = 460.8 m
(e) a = -10 ms-2
Monday, 15 March 2010
12. 2 (a)
vag = 20 ms-1
p = plane vpa = 60 ms-1
a = air
g = ground
vpg
vag
vpa
(b) We are told that the velocity of the plane relative to the ground needs to be in the
Westerly direction and asked what the pilot should do. In other words we are being asked
how the plane needs to moving on the moving frame of reference (the air). This is the
velocity of the plane relative to the air which is given by rearranging the equation (written
in 2(a)). p
v a = p
v g a
- v g
~ ~ ~
vpg = 60
θ (c) => tanθ = 20
-vag = 20
60
vpa
=> θ = 18.4o (S of E)
(d) vpa = 202 + 602
= 63.2 ms-1
Monday, 15 March 2010
13. 2 (a)
N vag = 20 ms-1
p = plane vpa = 60 ms-1
a = air
g = ground
vpg
vag
vpa
(b) We are told that the velocity of the plane relative to the ground needs to be in the
Westerly direction and asked what the pilot should do. In other words we are being asked
how the plane needs to moving on the moving frame of reference (the air). This is the
velocity of the plane relative to the air which is given by rearranging the equation (written
in 2(a)). p
v a = p
v g a
- v g
~ ~ ~
vpg = 60
θ (c) => tanθ = 20
-vag = 20
60
vpa
=> θ = 18.4o (S of E)
(d) vpa = 202 + 602
= 63.2 ms-1
Monday, 15 March 2010
14. 2 (a)
N vag = 20 ms-1
p = plane vpa = 60 ms-1
a = air
g = ground
vpg
vag
v gp
= p
v a + vag
~ ~ ~
vpa
(b) We are told that the velocity of the plane relative to the ground needs to be in the
Westerly direction and asked what the pilot should do. In other words we are being asked
how the plane needs to moving on the moving frame of reference (the air). This is the
velocity of the plane relative to the air which is given by rearranging the equation (written
in 2(a)). p
v a = p
v g a
- v g
~ ~ ~
vpg = 60
θ (c) => tanθ = 20
-vag = 20
60
vpa
=> θ = 18.4o (S of E)
(d) vpa = 202 + 602
= 63.2 ms-1
Monday, 15 March 2010
15. 3 (a)
vwg = 1.3
b = boat
w = water vbw = 2
g = ground vbg
(b) vbg = vbw + vwg
~ ~ ~
1.3
2
(c) vbg = 22 + 1.32
= 2.4 ms-1
(d) vwg = 1.3
tanθ = 1.3 => θ = tan-1(1.3/2) = 33o
2
vbw = 2
vbg => θ’ = 90 - 33
θ
= 57o
θ’
This is the angle relative to the bank
Monday, 15 March 2010
16. 3 (a)
vwg = 1.3
N
b = boat
w = water vbw = 2
g = ground vbg
(b) vbg = vbw + vwg
~ ~ ~
1.3
2
(c) vbg = 22 + 1.32
= 2.4 ms-1
(d) vwg = 1.3
tanθ = 1.3 => θ = tan-1(1.3/2) = 33o
2
vbw = 2
vbg => θ’ = 90 - 33
θ
= 57o
θ’
This is the angle relative to the bank
Monday, 15 March 2010
