Problem #1
An object's position during a given time interval is shown by the fig below:
1) positions A, B, C, D, and E
2) move from A to B, A to C, D to B and E to A.
(b) Answer the question (a) if point C is the reference point.
Answer:
(a) According to zero (reference):
1) position A, xA = –4 m; position B, xB = –1 m; position C, xC = + 2 m; position D, xD = +3 m; position E, xE = +5 m.
2) movement from A to B, ΔxAB = xB - xA = (–1 m) - (-4 m) = + 3 m (+ means right O)
the movement from A to C, ΔxAC = xC - xA = (+2 m) - (-4 m) = + 5 m (+ means right O)
displacement from D to B, ΔxDB = xB - xD = (–1 m) - (+3 m) = - 4 m ((-) means left O)
moving from E to A, xEA = xA - xE = (-4 m) - (+5 m) = - 9 m ((-) means left O)
(b) According to point C (reference):
1) position A, xA = –6 m; position B, xB = –3 m; position C, xC = 0 m; position D, xD = +1 m; position E, xE = + 3 m.
2) moving from A to B, ΔxAB = xB - xA = (-3 m) - (-6 m) = + 3 m (+ meaning right C)
the move from A to C, ΔxAC = xC - xA = (0) - (-6 m) = + 6 m (+ means right C)
displacement from D to B, ΔxDB = xB - xD = (-3 m) - (+1 m) = - 4 m ((-) means left C)
move from E to A, xEA = xA - xE = (-6 m) - (+3 m) = - 9 m ((-) means left C)
Problem #2
Find the following for the path in the figure below.
(a) the total distance traveled
(b) the displacement from start to finish
Answer:
(a) The distance is the total distance traveled regardless of direction:
8 m + 2 m + 3 m = 13 m.
(b) The displacement depends on the direction. Consider the right to be the positive direction and left to be negative. Then, add the displacements of each segment of the path:
8 m + (−2) m + 3 m = 9 m.
Or, since displacement is the change in position, simply find the final position with respect to the origin of the axis minus the initial position:
11 m − 2 m = 9 m.
Problem #3
A rock is thrown straight upward off the edge of a balcony that is 5 m above the ground. The rock rises 10 m, then falls all the way down to the ground below the balcony. What is the rock's displacement?
Answer:
Displacement deals only with the initial and final positions of the rock. The rock's initial position is the balcony, and its final position is the ground 5 m below. So, the rock's displacement is 5 m downward.
Problem #4
A child walks 5 m east, then 3 m north, then 1 m east.
a.) What is the magnitude of the child's displacement?
b.) What is the direction of the child's displacement?
Answer:
a.) The child's total distance is 5 + 3 + 1 = 9, however the displacement is the straight-line distance from the child's initial position to its final position. It is best to start by drawing a diagram.
(b) the displacement from start to finish
Answer:
(a) The distance is the total distance traveled regardless of direction:
8 m + 2 m + 3 m = 13 m.
(b) The displacement depends on the direction. Consider the right to be the positive direction and left to be negative. Then, add the displacements of each segment of the path:
8 m + (−2) m + 3 m = 9 m.
Or, since displacement is the change in position, simply find the final position with respect to the origin of the axis minus the initial position:
11 m − 2 m = 9 m.
Problem #3
A rock is thrown straight upward off the edge of a balcony that is 5 m above the ground. The rock rises 10 m, then falls all the way down to the ground below the balcony. What is the rock's displacement?
Answer:
Displacement deals only with the initial and final positions of the rock. The rock's initial position is the balcony, and its final position is the ground 5 m below. So, the rock's displacement is 5 m downward.
Problem #4
A child walks 5 m east, then 3 m north, then 1 m east.
a.) What is the magnitude of the child's displacement?
b.) What is the direction of the child's displacement?
Answer:
a.) The child's total distance is 5 + 3 + 1 = 9, however the displacement is the straight-line distance from the child's initial position to its final position. It is best to start by drawing a diagram.
As you can see in this diagram, the displacement is equivalent to the hypotenuse of a right triangle whose legs are 6 m and 3 m long. So, we can calculate the magnitude of the displacement using the Pythagorean theorem:
b.) The child's direction is equal to the angle formed between d and Δx, which we will call θ. Here we must use some trigonometry. So, we will use the fact that tanθ = opposite/adjacent to find θ:
tanθ = Δy/Δx
Problem #5
A particle moves 3m north then 4m east and finally 6m south.calculate the distance travelled and the displacement.
d = (Δx2 + Δy2)1/2 = (62 + 32)1/2 = 6.7 m
b.) The child's direction is equal to the angle formed between d and Δx, which we will call θ. Here we must use some trigonometry. So, we will use the fact that tanθ = opposite/adjacent to find θ:
tanθ = Δy/Δx
θ = tan-1(Δy/Δx) = tan-1(3/6) = tan-1(0.5) = 26.5o north of east
Notice the direction takes the form of an angle, 26.5o, and a reference for the angle, north of east. The second part is given so that it is clear what axis the angle is being measured from and in what direction from the axis. North of east means that the angle is measured from the east, or positive x, axis, and is north, the positive y direction, from this axis.Problem #5
A particle moves 3m north then 4m east and finally 6m south.calculate the distance travelled and the displacement.
Answer:
The particle starts from A and moves northwards to B. It then moves eastwards to C then southwards to D.
Hence, the total distance covered by the particle is
d = AB + BC + CD = 5 m + 6 m + 13 m = 24 m
Now, displacement is the shortest distance between the initial and the final positions. From the figure, it is clear that the displacement is AD.
From the figure, using phythagoras theorem
AD2 = AE2 + ED2 = (6 m)2 + (8 m)2 = 10 m
Hence, the displacement of the particle is 10 m
Problem #6
An athlete runs exactly once around a circular track with a total length of 500 m. Find the runner's displacement for the race.
AnswerSince the athlete runs exactly once around the circular track, that means that he/she finishes at the same point where he/she started, meaning the initial and final positions are the same. Thus, the displacement is zero, and the total length of the track is irrelevant to the question.
The particle starts from A and moves northwards to B. It then moves eastwards to C then southwards to D.
Hence, the total distance covered by the particle is
d = AB + BC + CD = 5 m + 6 m + 13 m = 24 m
Now, displacement is the shortest distance between the initial and the final positions. From the figure, it is clear that the displacement is AD.
From the figure, using phythagoras theorem
AD2 = AE2 + ED2 = (6 m)2 + (8 m)2 = 10 m
Hence, the displacement of the particle is 10 m
Problem #6
An athlete runs exactly once around a circular track with a total length of 500 m. Find the runner's displacement for the race.
AnswerSince the athlete runs exactly once around the circular track, that means that he/she finishes at the same point where he/she started, meaning the initial and final positions are the same. Thus, the displacement is zero, and the total length of the track is irrelevant to the question.
Post a Comment for "Displacement and Distance Problem and Solution"