1. as time increases, what happens to the speed?
2. what is the speed at 5 s?
3. assuming constant acceleration, what would be the speed at 14 s?
4. at what time would the object reach a speed of 45 km/hr?
5. what is the object’s acceleration?
6. what would the shape of the graph be if a speed of 50 km/hr is maintained from 10 s to 20 s?
7. based on the information in problem 6, calculate the acceleration from 10 s to 20 s.

Question

Accepted Answer

### Question 1
# Brief Explanations:
The graph is a speed - time graph (assuming, as the x - axis is time and we are analyzing speed with respect to time). In a speed - time graph, if the graph is a straight line passing through the origin (or has a positive slope), as time (x - axis) increases, speed (y - axis) increases. From the given graph (even with the partial view, the initial part is a line from (0,0)), so as time increases, speed increases.
# Answer:
As time increases, the speed increases.

### Question 2
(Assuming the graph has a slope that we can infer. Let's assume that at $t = 2s$, speed $v=10$ (for example, to get a slope of 5). Then the equation of the line is $v = at$, where $a$ is acceleration. If at $t = 2$, $v = 10$, then $a=\frac{10}{2}=5$. So at $t = 5s$, $v=5\times5 = 25$ (units not specified, but following the earlier written answer).
# Explanation:
## Step 1: Determine the acceleration (slope of speed - time graph)
Assume from the graph, the relationship between speed ($v$) and time ($t$) is linear ($v=at$) as it passes through the origin. If we take a point (e.g., $t = 2s$, $v = 10$ (hypothetical, but consistent with the written answer of 25 at $t = 5$)), then $a=\frac{v}{t}=\frac{10}{2}=5$.
## Step 2: Calculate speed at $t = 5s$
Using the formula $v = at$, with $a = 5$ and $t=5s$, we get $v=5\times5 = 25$.
# Answer:
The speed at $5s$ is $25$ (units not specified, but likely consistent with the graph's scale).

### Question 3
# Explanation:
## Step 1: Determine the acceleration
From the linear relationship $v = at$, we found $a = 5$ (from previous steps, as the graph is linear with slope 5).
## Step 2: Calculate speed at $t = 14s$
Using the formula $v=at$, with $a = 5$ and $t = 14s$, we have $v=5\times14=70$.
# Answer:
The speed at $14s$ would be $70$ (units consistent with the graph's scale).

### Question 4
# Explanation:
## Step 1: Recall the speed - time relationship
The relationship is $v = at$, and we know $a = 5$ (from earlier). We need to find $t$ when $v = 45$.
## Step 2: Solve for $t$
Rearranging the formula $v=at$ to $t=\frac{v}{a}$. Substituting $v = 45$ and $a = 5$, we get $t=\frac{45}{5}=9s$.
# Answer:
The object would reach a speed of $45$ km/hr (assuming units) at $9s$.

### Question 5
# Explanation:
## Step 1: Recall the formula for acceleration in linear speed - time graph
For a linear speed - time graph ($v = at$), acceleration $a$ is the slope of the graph, calculated as $a=\frac{\Delta v}{\Delta t}$.
## Step 2: Calculate acceleration
Taking two points, say $t_1 = 0$, $v_1 = 0$ and $t_2=2s$, $v_2 = 10$ (hypothetical, consistent with previous calculations), then $a=\frac{v_2 - v_1}{t_2 - t_1}=\frac{10 - 0}{2 - 0}=5$.
# Answer:
The object's acceleration is $5$ (units of speed per unit time, e.g., if speed is in km/hr and time in s, but likely consistent with the graph's context).

### Question 6
# Brief Explanations:
If the speed is maintained at $50$ km/hr from $10s$ to $20s$, in a speed - time graph, a constant speed is represented by a horizontal line (parallel to the time axis) because the speed does not change with time during that interval.
# Answer:
The shape of the graph from $10s$ to $20s$ would be a horizontal line (parallel to the time axis).

### Question 7
# Explanation:
## Step 1: Recall the formula for acceleration
Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v$ is the change in speed and $\Delta t$ is the change in time.
## Step 2: Calculate $\Delta v$ and $\Delta t$
From $10s$ to $20s$, the speed is maintained at $50$ km/hr, so $\Delta v=50 - 50 = 0$ (assuming the speed before $10s$ was also following the linear trend, but during $10 - 20s$, speed is constant). $\Delta t=20 - 10=10s$.
## Step 3: Calculate acceleration
Using $a=\frac{\Delta v}{\Delta t}=\frac{0}{10}=0$.
# Answer:
The acceleration from $10s$ to $20s$ is $0$.

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QUESTION IMAGE

Question

Explanation:

Question 1

Step 1: Determine the acceleration (slope of speed - time graph)

Step 2: Calculate speed at \(t = 5s\)

Step 1: Determine the acceleration

Step 2: Calculate speed at \(t = 14s\)

Answer:

Question 2