When running a 100-meter race Henry reaches his maximum speed when he is 22 meters from the starting line and 4 seconds have elapsed since the start of the race. Henry maintains this maximum (constant) speed for the rest of the race. If Henry is 43 meters from the starting line 7 seconds after the start of the race:



Let's break down the problem step by step.

### Information provided:
- Henry reaches his maximum speed when he is 22 meters from the starting line and 4 seconds have elapsed since the start.
- After that, Henry maintains his maximum speed for the rest of the race.
- Henry is 43 meters from the starting line 7 seconds after the start of the race.

We can solve the problem by calculating Henry's maximum speed and then modeling his movement after he reaches it.

### Step 1: Maximum Speed Calculation
Henry maintains a constant speed after 4 seconds, and he travels from 22 meters (at 4 seconds) to 43 meters (at 7 seconds), which is a distance of:

\[
\Delta x = 43 \, \text{meters} - 22 \, \text{meters} = 21 \, \text{meters}
\]

The time elapsed between 4 seconds and 7 seconds is:

\[
\Delta t = 7 \, \text{seconds} - 4 \, \text{seconds} = 3 \, \text{seconds}
\]

Thus, Henry's maximum (constant) speed after 4 seconds is:

\[
\text{Maximum speed} = \frac{\Delta x}{\Delta t} = \frac{21 \, \text{meters}}{3 \, \text{seconds}} = 7 \, \text{meters per second}
\]

### Step 2: Distance Covered Before Reaching Maximum Speed
Now, we know Henry reaches 22 meters at 4 seconds. Let’s assume he accelerated uniformly from the start to 4 seconds, then reached this constant speed. We can use the following kinematic equation for the distance covered under constant acceleration:

\[
x(t) = \frac{1}{2} a t^2
\]

Where:
- \( x(t) = 22 \, \text{meters} \)
- \( t = 4 \, \text{seconds} \)

Substituting into the equation:

\[
22 = \frac{1}{2} a (4)^2
\]

Solving for \( a \) (acceleration):

\[
22 = 8a \quad \Rightarrow \quad a = \frac{22}{8} = 2.75 \, \text{meters per second squared}
\]

### Step 3: Speed at 4 Seconds
The speed at 4 seconds can be calculated using the equation:

\[
v(t) = a t
\]

Substituting \( a = 2.75 \, \text{m/s}^2 \) and \( t = 4 \, \text{seconds} \):

\[
v(4) = 2.75 \times 4 = 11 \, \text{meters per second}
\]

However, since we've already calculated that his constant speed after 4 seconds is \( 7 \, \text{m/s} \), this value would have been reached at some point before 4 seconds, and the remaining acceleration from 22m to 43m is under constant speed.

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When running a 100-meter race Henry reaches his maximum speed when he is 22 meters from the starting line and 4 seconds have elapsed since the start of the race. Henry maintains this maximum (constant) speed for the rest of the race. If Henry is 43 meters from the starting line 7 seconds after the start of the race:

Discover how to calculate Henry's maximum speed during a 100-meter race using kinematic equations and motion principles. This problem involves constant speed and uniform acceleration concepts and is suitable for Grades 9-12.

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