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# Questions about velocity, acceleration & force

(See attached file for full problem description)

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As a vehicle goes from +4m/s to -1 m/s, what is its change in velocity?

For 2 and 3 use the following:
Vector P: 50 meters @ 110 degrees
Vector Q: 35 meters @ 315 degrees

Question 2:
What is 3P - 3Q?

Question 3:
What is 2P + 3Q?

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?

Question 5:
A velocity vector of 24 m/s is at an angle of 320 degrees. What is its vertical component?

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.

Question 7:
Vector lengths of 30m, 500m and X create a resultant length of 10m. What could X be?

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)

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.

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.

Question 11:
Determine the final velocity of the wagon after a time of .3996 seconds if it starts from rest.

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.

Question 12:
Determine the magnitude of the net force provided by these two forces.

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.

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)
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#### Solution Preview

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.
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For 2 and 3 use the following:
Vector P: 50 meters @ 110 degrees
Vector Q: 35 meters @ 315 degrees
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 ...

#### Solution Summary

The four questions related to Mechanics about velocity acceleration and force are solved and explained.

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