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Question 1:
As a vehicle goes from +4m/s to -1 m/s, what is its change in velocity?
Answer:
As velocity is a vector quantity the direction is required to express it completely.
Here the directions are given in terms of +ve and –ve sign, it means that the vehicle is moving
in a straight line hence forward velocity is taken +ve and backward as –ve. The change in
velocity is final velocity – initial velocity hence
v
= v2 – v1 = (-1) – (+4) = - 5 m/s.
--------------------------------------------
Y
For 2 and 3 use the following:
P
Vector P: 50 meters @ 110 degrees
Vector Q: 35 meters @ 315 degrees

x
Question 2:
What is 3P - 3Q?
Answer:
If the vectors are resolved as components in x and y direction then we can add or
subtract the components independently. The process is as follows
(Bold letter is for vector, P is magnitude of P, .i and j are unit vectors in x and y directions
respectively.)
P = P cos i + P sin  j = 50 cos 1100 i + 50 sin 1100 j = - 17.1 i + 47.0 j and similarly
Q = 35 cos 3150 i + 35 sin 3150 j = 24.7 i- 24.7 j
Hence 3P –3Q = 3(P – Q) = 3[(-17i + 47j) – (24.7i – 24.7j)] = 3(- 41.7i + 71.7j)
= - 125.1i + 215.1j
As the magnitude of A =Axi + Ayj is given by A = (Ax2 + Ay2) and direction by tan-1(Ay/Ax)
magnitude of 3P –3Q is [(- 125.1)2 + (215.1)2] =(15650 + 46268) = 248.8 m, and the
direction will be tan-1(215.1/-125.1) = tan-1(-1.7) = -59.80 or 1800 -59.80 =0120.20. As P is in
second quadrant and Q is in forth quadrant so – Q is also in second quadrant, hence the
resultant of P and –Q must be in second quadrant and the correct value of the angle is 120.20.
Q2Ansa: 248.8 meters Q2Ansb: 120.2 degrees
Question 3:
What is 2P + 3Q?
Answer: As above question
2P + 3Q = 2(-17i + 47j) + 3(24.7i – 24.7j) =(- 34i + 94j) + (74.1i – 74.1j) = 40.1i + 19.9j
hence magnitude of 2P + 3Q will be (40.12 + 19.92) = 44.8 m, and the direction is given by
tan-1(19.9/40.1) = 26.40 or 153.60. As P is in second and Q is in forth quadrant as well as P is at
1100 the correct value is 26.40.
Q3Ansa: 44.8 meters Q3Ansb: 26.4 degrees
Question 4:
Dick Rutan, the man attempting to fly around the world in a balloon, is floating due east at 18
km/hr. One hour later he has turned and is now sailing due south at 48 km/hr. What is the
magnitude of his average acceleration over this period of time?
Answer:
Writing in the components form with east as x and north as y directions we have
V1 = 18i + 0j and V2 = 0i – 48j hence the change in velocity is V2 – V1 = -18i – 48j whose
magnitude is [(-18)2 +(-48)2] = 51.3 km/hr. As this change in velocity is in one hour the
acceleration is 51.3/1 =51.3 km/hr/hr.
Q4Ans:
51.3 km/hr/hr
Question 5:
A velocity vector of 24 m/s is at an angle of 320 degrees. What is its vertical component?
Q5Ans: The angle given appears to be with horizontal hence the vertical component should be
24 sin = 24 sin 320 = - 15.4 m/s (-ve shows downward)
- 15.4 m/s
Question 6:
A 25 kg mass starts from rest and is forced to the right with a force of 14.98 Newtons and has a
force of friction of 7.49 Newtons. Determine the distance the mass travels in 19.98 seconds.
Q6Ans: The acceleration a is given by F – Ff =ma or a = (F – Ff) / m = (14.98 – 7.49) / 25
= 3 m/s/s, hence the distance traveled is given by
x = ut + (0.5)at2
= 0 + (0.5)x 3 x 19.982 = 598.80 m
598.80meters
Question 7:
Vector lengths of 30m, 500m and X create a resultant length of 10m. What could X be?
Q7Ans:
As the directions of the vectors are not given we have to surch for max and min
possible values of x. The resultant of 500m and 30m is having maximum of 530m when they in
same direction and minimum of 470m if they are in apposite directions. When we find resultant
10m with 530 and x value of x should be 520m or 540m and with 470m it should be 460m and
480m. Hence x could be a minimum of 460m and a maximum value of 540m.
Between 460m and 540 meters
Question 8:
Six vectors are shown in the diagram at right.
What is a possible direction on the resultant
vector of 1 + 2 + 3 + 4 + 6? (Use only positive
values)
Q8Ans: I think that the magnitudes of the vectors are shown on the graph. Writing the vectors
in component form and adding we get
1+2+3+4+5+6 = (-3i -3j)+(-10i)+(6i-7j)+(7i+4j)+(5i+2j)+(-2j) = 5i –6j
hence direction of resultant is tan-1(-6/5) = -50.190
-50.19 degrees
Question 9:
A 15.98 kg sky diver falls and experiences an upward force due to air resistance of 10.99 N at a
point in time. Determine the magnitude of the acceleration the sky diver experiences at this
time.
The acceleration is given by the equation mg – F = ma or a = (mg – F)/m
= (15.98 x 9.8 – 10.99)/15.98 = 9.11 m/s/s (downward)
Q9Ans:
9.11 m/s/s downwards.
Use the following situation to answer questions 10 & 11. A 10.99 kg wagon is pulled to the
right with a 15.98 N force that is applied at an angle of 30 degrees with respect to the
horizontal. There is a 4.99 Newton force of friction pulling the wagon to the left.
Question 10:
Determine the acceleration of the wagon.
Q10Ans:
The horizontal component of the force applied is 15.98 cos 30 =13.84 N
hence the net force on the wagon 13.84 – 4.99 = 8.85 N, therefore the acceleration of the wagon
is 8.85/10.99 = 0.805 m/s/s
0.805 m/s/s
Question 11:
Determine the final velocity of the wagon after a time of .3996 seconds if it starts from rest.
Q11Ans:
Using v = u + at = 0 + 0.805 x 0.3996 = 0.322 m/s
0.322 m/s
For questions 12 - 14 below use the following situation: Two forces act concurrently on a 25 kg
mass: a 69.95 Newton force at 45 degrees along with a 98.92 Newton force at 120 degrees.
With the same notations used above (tri to do it directly)
F1 + F2 = 69.95 (cos 450i + sin450j) + 98.92(cos1200i + sin1200j)
= (49.46 – 49.46)i + (49.46 + 85.67)j = 0i +135.13j
Question 12:
Determine the magnitude of the net force provided by these two forces.
Q12Ans: The magnitude of the net force 0i +135.13j is 135.13N
135.13 Newtons
Question 13:
Determine the force that would need to act along with these two forces to put a zero net force
on the 25 kg mass.
Q13Ans:
135.13 Newtons
Question 14:
Determine the direction at which the force in Question 13 would need to act in order to put a
zero net force on the 25 kg mass. (use 0 degrees east as reference)
Q14Ans: To make the resultant zero the third force must be – 135.13j N means it must be in
negative y direction
270 degrees
PLEASE!!!! Do not put such large number of questions in one posting.