Home Page Textbooks Physics Page 47: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 47: Lesson Check

Exercise 3

Solution 1

Solution 2

Step 1

1 of 3

$textbf{(a)}$ Distance is the total length travelled, and it is defined as always being positive. This means that $textbf{the distance on a round-trip is positive,}$ regardless of the fact that the initial and final positions are the same.

Step 2

2 of 3

$textbf{(b)}$ In contrast with distance, displacement is the net change in position. This means that $textbf{the displacement on a round-trip is zero,}$ the reason being that the initial and final positions are the same, so there’s no net change in position.

Result

3 of 3

$textbf{(a)}$ The distance on a round-trip is positive.

$textbf{(b)}$ The displacement on a round-trip is zero.

Step 1

1 of 3

Distance is a scalar quantity that measures the total length of the path taken without taking into account any changes in direction.

Step 2

2 of 3

Displacement is a vector quantity that is described by both distance and direction.

If the trip is one way without changing direction, then the magnitude of distance is the same as the magnitude of displacement.

Result

3 of 3

(a) The distance traveled will always be positive for the entire trip, either round trip or one way.
(b) The displacement for a round trip will always be zero because the initial and final positions are the same.

Exercise 4

Solution 1

Solution 2

Step 1

1 of 2

$textbf{An odometer in a car measures distance.}$ A very simplified explanation to how an odometer works is by counting the number wheel rotations during your trip, and assuming that the distance travelled is the number of wheel rotations multiplied with the tire circumference.

Result

2 of 2

An odometer in a car measures distance.

Step 1

1 of 2

A GPS device used in some cars can tell your position therefore giving your exact location. Some software uses GPS to map your motion and can give information such as displacement and velocity.

Result

2 of 2

The odometer just measures total distance traveled and cannot tell the direction so it can’t measure displacement.

Exercise 5

Solution 1

Solution 2

Step 1

1 of 2

$textbf{The answer is yes,}$ you can take a hike and have the distance covered be equal to the magnitude of your displacement. $textbf{That’s possible only if you walk in a straight line without changing the direction.}$ If you deviate from a straight line or change direction, the distance you’ve covered and the magnitude of the displacement are no longer equal.

Result

2 of 2

Yes; the distance you cover is equal to the magnitude of your displacement if you take a hike in a straight line without changing the direction.

Step 1

1 of 2

Hiking in a straight line means that the total distance and the displacement will have the same magnitude.This might change though if you go up hill since you are changing your vertical direction.

Result

2 of 2

Yes. Distance traveled and the magnitude of the displacement are equal if you walk in a straight line in one direction.

Exercise 6

Solution 1

Solution 2

Step 1

1 of 3

$textbf{(a)}$ $textbf{When you arrive at the park, you and your dog have the same displacement.}$ That’s because both you and your dog have the same initial and final positions.

Step 2

2 of 3

$textbf{(b)}$ $textbf{The distance you and your dog have travelled is not the same.}$ Because of the many short side trips, your dog will have covered more distance than you.

Result

3 of 3

$textbf{(a)}$ When arriving at the park, you and your dog have the same displacement because the initial and final positions for both of you are the same.

$textbf{(b)}$ You and your dog haven’t travelled the same distance because of the side trips your dog took while walking to the park.

Step 1

1 of 3

a.) $text{color{#4257b2} boxed{bf Yes}, since both you and the dog had the same initial and the final position so the displacement is same for you and the dog}$

Step 2

2 of 3

b.) $text{color{#4257b2} boxed{bf No}, since the dog zigzags around as it takes many short side trips to chase squirrels, examine fire hydrants, and so on so the distance travelled by the dog and you may be different.}$

Result

3 of 3

Click here to see the explanation.

Exercise 7

Step 1

1 of 3

$textbf{(a)}$ The total distance travelled by the ball is the sum of all traversed distances. In this case, the ball is initially $5.0 text{m}$ from the hole, then the golfer overshoots the hole by $1.2 text{m}$, and lastly the golfer manages to hit the golf ball into the hole from that position. Therefore, the distance travelled by the ball is

$$
begin{align*}
d&=5.0 text{m}+1.2 text{m}+1.2 text{m}\
&=quadboxed{7.4 text{m}}\
end{align*}
$$

Step 2

2 of 3

$textbf{(b)}$ The displacement of the ball is the net change in its position. Setting the origin of the coordinate system at the initial point, $x_{i}=0.0 text{m}$, we get that the displacement of the ball is

$$
begin{align*}
Delta x&=x_{f}-x_{i}\
&=5.0 text{m}-0.0 text{m}\
&=quadboxed{5.0 text{m}}\
end{align*}
$$

Result

3 of 3

$textbf{(a)}$ $boxed{d=7.4 text{m}}$

$textbf{(b)}$ $boxed{Delta x=5.0 text{m}}$

Exercise 8

Step 1

1 of 3

$textbf{(a)}$ The distance covered by the ball is the sum of the distances travelled in the positive and negative direction with respect to the initial position. It travels $22 text{cm}$ in the positive direction, and comes to rest $7.5 text{cm}$ behind its original position. Therefore, the total distance covered by the ball is

$$
begin{align*}
d&=22 text{cm}+22 text{cm}+7.5 text{cm}\
&=quadboxed{51.5 text{cm}}\
end{align*}
$$

Step 2

2 of 3

$textbf{(b)}$ The initial position of the ball is at $x_{i}=0 text{cm}$, whereas the final position is $7.5 text{cm}$ $textit{behind}$ the original position. Therefore, the displacement of the ball is

$$
begin{align*}
Delta x&=x_{f}-x_{i}\
&=-7.5 text{cm}-0 text{cm}\
&=quadboxed{-7.5 text{cm}}\
end{align*}
$$

Result

3 of 3

$textbf{(a)}$ $boxed{d=51.5 text{cm}}$

$textbf{(b)}$ $boxed{Delta x=-7.5 text{cm}}$

Exercise 9

Solution 1

Solution 2

Step 1

1 of 4

$textbf{(a)}$ $textbf{The distance covered by the train is greater than its displacement.}$ Distance is a scalar quantity, and is defined as always being positive, i.e., the sum of all individual distances covered during motion. It is not associated with direction in any way. On the other hand, displacement is a vector, which means that it’s important to consider the direction of motion as well. All that being said, one can conclude that $textbf{displacement can never be greater than total distance.}$

Step 2

2 of 4

$textbf{(b)}$ The distance covered by the train is the sum of the distances travelled in positive and negative direction,

$$
begin{align*}
d&=5.9 text{km}+3.8 text{km}\
&=quadboxed{9.7 text{km}}\
end{align*}
$$

Step 3

3 of 4

$textbf{(c)}$ First, let’s set the origin at $x=0.0 text{km}$, which will also be the train’s initial position; $x_{i}=0 text{km}$. From there, the train travels for $5.9 text{km}$ in the positive direction, and then backs up for $3.8 text{km}$. This means that the final position of the train is

$$
begin{align*}
x_{f}&=5.9 text{km}-3.8 text{km}\
&=2.1 text{km}
end{align*}
$$

from the initial position. Therefore, the displacement of the train is

$$
begin{align*}
Delta x&=x_{f}-x{i}\
&=2.1 text{km}-0 text{km}\
&=quadboxed{2.1 text{km}}\
end{align*}
$$

Result

4 of 4

$textbf{(a)}$ The distance covered by the train is greater than its displacement.

$textbf{(b)}$ $boxed{d=9.7 text{km}}$

$textbf{(c)}$ $boxed{Delta x=2.1 text{km}}$

Step 1

1 of 4

$$
tt{(a) Distance is a scalar quantity and it is always positive, it does not have a direction, whereas displacement is a vector quantity and by definition $Delta x =x_f-x_i$, the displacement has a direction and can be positive, negative or zero, when comparing distance and displacement we can say that displacement can never be greater than the distance covered, $Delta x_{max} le d$\
as a conclusion, we can say the distance covered by the train is{ color{#4257b2}{greater than}} its displacement. }
$$

Step 2

2 of 4

$$
tt{(b) distance covered: $5.9+3.8=boxed{9.7km}$ }
$$

Step 3

3 of 4

$$
tt{(c) train displacement : $5.9-3.8=boxed{2.1km}$}
$$

Result

4 of 4

tt{(a) Greater then,(b) 9.7km,(c) 2.1km}

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