motion calculations
1. jane traveled from gaithersburg high school to new york by the quickest route. google maps says this trip should take 3.72 hours. what would her average speed be if google maps is right about the time?
2. if the car accelerates from 0 to 40 mph in 5 seconds, what is its acceleration?
3. if a person runs across a street at 10 mph then hits a pole and stops in ¼ second, what is their acceleration?
4. a football player catches the football at his own 30 yard line. forward is defined as the positive direction. if he runs with a velocity of +5 yards per second for 6 seconds, where will he be when he stops?
5. let’s say the football player starts at the 30 yard line, runs backward 10 yards in 4 seconds and then forward 30 yards in the following 6 seconds. what is his average velocity?
6. use this graph to find the velocity from…
a. …1 to 3 seconds
b. …3 to 4.5 seconds

Question

motion calculations
1. jane traveled from gaithersburg high school to new york by the quickest route. google maps says this trip should take 3.72 hours. what would her average speed be if google maps is right about the time?
2. if the car accelerates from 0 to 40 mph in 5 seconds, what is its acceleration?
3. if a person runs across a street at 10 mph then hits a pole and stops in ¼ second, what is their acceleration?
4. a football player catches the football at his own 30 yard line. forward is defined as the positive direction. if he runs with a velocity of +5 yards per second for 6 seconds, where will he be when he stops?
5. let’s say the football player starts at the 30 yard line, runs backward 10 yards in 4 seconds and then forward 30 yards in the following 6 seconds. what is his average velocity?
6. use this graph to find the velocity from…
  a. …1 to 3 seconds
  b. …3 to 4.5 seconds

Accepted Answer

### Problem 1: Jane's Average Speed
# Explanation:
## Step1: Recall the formula for average speed
The formula for average speed $ v $ is $ v=\frac{d}{t} $, where $ d $ is the distance and $ t $ is the time. But we need to know the distance. From the map, the distance (quickest route) - let's assume the distance from Gaithersburg High School to New York via the quickest route is, for example, if we consider the time and we need to find speed, but maybe the distance is missing? Wait, maybe the map shows the distance? Wait, the user's image: the map has "4 h 33 min" and "3 h 42 min" - maybe the distance is, let's check typical distance from Gaithersburg (MD) to New York: about 240 miles (approximate). Let's assume the distance $ d = 240 $ miles (since it's a common distance). Time $ t=3.72 $ hours.

## Step2: Calculate average speed
Using $ v=\frac{d}{t} $, substitute $ d = 240 $ miles and $ t = 3.72 $ hours.
$ v=\frac{240}{3.72}\approx64.52 $ miles per hour.

# Answer:
Approximately $ 64.52 $ mph (assuming distance is 240 miles; if distance is different, recalculate with actual distance).

### Problem 2: Car's Acceleration
# Explanation:
## Step1: Recall acceleration formula
Acceleration $ a=\frac{\Delta v}{\Delta t} $, where $ \Delta v $ is change in velocity and $ \Delta t $ is change in time.

## Step2: Identify values
Initial velocity $ v_i = 0 $ mph, final velocity $ v_f = 40 $ mph, time $ \Delta t = 5 $ seconds. First, convert time to hours or velocity to miles per second? Let's convert 5 seconds to hours: $ 5 $ seconds $=\frac{5}{3600}$ hours $=\frac{1}{720}$ hours. Or convert velocity to miles per second: $ 40 $ mph $=\frac{40}{3600}$ miles per second $=\frac{1}{90}$ miles per second.

Using $ a=\frac{v_f - v_i}{\Delta t} $, with $ v_f - v_i = 40 - 0 = 40 $ mph, $ \Delta t = 5 $ seconds $=\frac{5}{3600}$ hours. So $ a=\frac{40}{\frac{5}{3600}} = 40\times\frac{3600}{5}= 28800 $ miles per hour squared. Or in miles per second squared: $ \Delta v = \frac{40}{3600} $ miles per second, $ \Delta t = 5 $ seconds, so $ a=\frac{\frac{40}{3600}}{5}=\frac{40}{3600\times5}=\frac{40}{18000}=\frac{1}{450}\approx0.00222 $ miles per second squared. Alternatively, convert to ft/s²: 40 mph = $ 40\times\frac{5280}{3600}=\frac{211200}{3600}\approx58.6667 $ ft/s. Time = 5 s. So $ a=\frac{58.6667 - 0}{5}\approx11.73 $ ft/s².

# Answer:
If using mph and hours: $ 28800 $ mph²; if using ft/s and s: $ \approx11.73 $ ft/s².

### Problem 3: Person's Acceleration
# Explanation:
## Step1: Acceleration formula
$ a=\frac{\Delta v}{\Delta t} $, where $ \Delta v = v_f - v_i $, $ v_i = 10 $ mph, $ v_f = 0 $ mph, $ \Delta t=\frac{1}{4} $ second.

## Step2: Convert units
First, convert 10 mph to ft/s: $ 10\times\frac{5280}{3600}=\frac{52800}{3600}\approx14.6667 $ ft/s. $ \Delta t = 0.25 $ s. $ \Delta v = 0 - 14.6667=- 14.6667 $ ft/s.

## Step3: Calculate acceleration
$ a=\frac{-14.6667}{0.25}=-58.6668 $ ft/s² (negative because it's deceleration).

# Answer:
$ - 58.67 $ ft/s² (or in other units, e.g., mph/s: $ \frac{0 - 10}{\frac{1}{4}}=-40 $ mph/s).

### Problem 4: Football Player's Position
# Explanation:
## Step1: Recall position formula
For constant velocity, $ x = x_0 + vt $, where $ x_0 $ is initial position, $ v $ is velocity, $ t $ is time.

## Step2: Identify values
$ x_0 = 30 $ yard line, $ v = + 5 $ yards per second, $ t = 6 $ seconds.

## Step3: Calculate final position
$ x=30+(5\times6)=30 + 30 = 60 $ yard line.

# Answer:
60 yard line.

### Problem 5: Football Player's Average Velocity
# Explanation:
## Step1: Average velocity formula
Average velocity $ \bar{v}=\frac{\Delta x}{\Delta t} $, where $ \Delta x $ is displacement, $ \Delta t $ is total time.

## Step2: Calculate displacement
Starts at 30 yard line, runs backward 10 yards (displacement $ - 10 $ yards) then forward 30 yards (displacement $ + 30 $ yards). Wait, the problem says: "starts at the 30 yard line, runs backward 10 yards in 4 seconds and then forward 30 yards in the following 6 seconds". So total displacement $ \Delta x=( - 10)+30 = 20 $ yards (from initial position: 30 - 10 + 30 = 50? Wait, no: initial position $ x_0 = 30 $. After backward: $ x_1=30 - 10 = 20 $. After forward: $ x_2=20 + 30 = 50 $. So displacement $ \Delta x=50 - 30 = 20 $ yards. Total time $ \Delta t=4 + 6 = 10 $ seconds.

## Step3: Calculate average velocity
$ \bar{v}=\frac{20}{10}=2 $ yards per second.

# Answer:
2 yards per second.

### Problem 6a: Velocity from 1 to 3 seconds (Position - Time Graph)
# Explanation:
## Step1: Velocity from position - time graph
Velocity $ v=\frac{\Delta x}{\Delta t} $, where $ \Delta x $ is change in position, $ \Delta t $ is change in time.

## Step2: Identify positions at t = 1 and t = 3
From the graph: at $ t = 1 $ s, position $ x_1 = 20 $ m. At $ t = 3 $ s, position $ x_2 = 60 $ m.

## Step3: Calculate velocity
$ \Delta x=60 - 20 = 40 $ m, $ \Delta t=3 - 1 = 2 $ s. So $ v=\frac{40}{2}=20 $ m/s.

# Answer:
20 m/s.

### Problem 6b: Velocity from 3 to 4.5 seconds (Position - Time Graph)
# Explanation:
## Step1: Velocity from position - time graph
$ v=\frac{\Delta x}{\Delta t} $.

## Step2: Identify positions at t = 3 and t = 4.5
From the graph: at $ t = 3 $ s, position $ x_1 = 60 $ m. At $ t = 4.5 $ s, position $ x_2 = 60 $ m (since the graph is flat there).

## Step3: Calculate velocity
$ \Delta x=60 - 60 = 0 $ m, $ \Delta t=4.5 - 3 = 1.5 $ s. So $ v=\frac{0}{1.5}=0 $ m/s.

# Answer:
0 m/s.

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QUESTION IMAGE

Question

Explanation:

Problem 1: Jane's Average Speed

Step1: Recall the formula for average speed

Step2: Calculate average speed

Step1: Recall acceleration formula

Step2: Identify values

Step1: Acceleration formula

Step2: Convert units

Step3: Calculate acceleration

Answer:

Problem 2: Car's Acceleration