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Sagot :
Sure, let's solve this problem step-by-step. The given equation is:
[tex]\[ t = \sqrt{x} + 3 \][/tex]
### Step 1: Isolate [tex]\( \sqrt{x} \)[/tex]
Subtract 3 from both sides of the equation:
[tex]\[ t - 3 = \sqrt{x} \][/tex]
### Step 2: Square Both Sides
To get rid of the square root, we need to square both sides of the equation:
[tex]\[ (t - 3)^2 = \sqrt{x}^2 \][/tex]
Since [tex]\(\sqrt{x}\)[/tex] squared is [tex]\(x\)[/tex], we now have:
[tex]\[ (t - 3)^2 = x \][/tex]
So, the displacement [tex]\(x\)[/tex] in terms of time [tex]\(t\)[/tex] is:
[tex]\[ x = (t - 3)^2 \][/tex]
### Step 3: Calculate Displacement for Various Time Values
To understand how the body moves over time, let's calculate the displacement [tex]\(x\)[/tex] for different values of [tex]\(t\)[/tex]. Here are the results for [tex]\(t\)[/tex] values from 1 to 10:
- For [tex]\(t = 1\)[/tex]:
[tex]\[ x = (1 - 3)^2 = (-2)^2 = 4 \][/tex]
- For [tex]\(t = 2\)[/tex]:
[tex]\[ x = (2 - 3)^2 = (-1)^2 = 1 \][/tex]
- For [tex]\(t = 3\)[/tex]:
[tex]\[ x = (3 - 3)^2 = 0^2 = 0 \][/tex]
- For [tex]\(t = 4\)[/tex]:
[tex]\[ x = (4 - 3)^2 = 1^2 = 1 \][/tex]
- For [tex]\(t = 5\)[/tex]:
[tex]\[ x = (5 - 3)^2 = 2^2 = 4 \][/tex]
- For [tex]\(t = 6\)[/tex]:
[tex]\[ x = (6 - 3)^2 = 3^2 = 9 \][/tex]
- For [tex]\(t = 7\)[/tex]:
[tex]\[ x = (7 - 3)^2 = 4^2 = 16 \][/tex]
- For [tex]\(t = 8\)[/tex]:
[tex]\[ x = (8 - 3)^2 = 5^2 = 25 \][/tex]
- For [tex]\(t = 9\)[/tex]:
[tex]\[ x = (9 - 3)^2 = 6^2 = 36 \][/tex]
- For [tex]\(t = 10\)[/tex]:
[tex]\[ x = (10 - 3)^2 = 7^2 = 49 \][/tex]
### Summary
Here are the displacements [tex]\(x\)[/tex] for each time [tex]\(t\)[/tex]:
- [tex]\(t = 1 \Rightarrow x = 4\)[/tex]
- [tex]\(t = 2 \Rightarrow x = 1\)[/tex]
- [tex]\(t = 3 \Rightarrow x = 0\)[/tex]
- [tex]\(t = 4 \Rightarrow x = 1\)[/tex]
- [tex]\(t = 5 \Rightarrow x = 4\)[/tex]
- [tex]\(t = 6 \Rightarrow x = 9\)[/tex]
- [tex]\(t = 7 \Rightarrow x = 16\)[/tex]
- [tex]\(t = 8 \Rightarrow x = 25\)[/tex]
- [tex]\(t = 9 \Rightarrow x = 36\)[/tex]
- [tex]\(t = 10 \Rightarrow x = 49\)[/tex]
This calculation shows how the displacement [tex]\(x\)[/tex] varies over time [tex]\(t\)[/tex]. The body starts with a displacement of 4 units at [tex]\(t = 1\)[/tex], decreases to 0 at [tex]\(t = 3\)[/tex], and then increases again as time progresses.
[tex]\[ t = \sqrt{x} + 3 \][/tex]
### Step 1: Isolate [tex]\( \sqrt{x} \)[/tex]
Subtract 3 from both sides of the equation:
[tex]\[ t - 3 = \sqrt{x} \][/tex]
### Step 2: Square Both Sides
To get rid of the square root, we need to square both sides of the equation:
[tex]\[ (t - 3)^2 = \sqrt{x}^2 \][/tex]
Since [tex]\(\sqrt{x}\)[/tex] squared is [tex]\(x\)[/tex], we now have:
[tex]\[ (t - 3)^2 = x \][/tex]
So, the displacement [tex]\(x\)[/tex] in terms of time [tex]\(t\)[/tex] is:
[tex]\[ x = (t - 3)^2 \][/tex]
### Step 3: Calculate Displacement for Various Time Values
To understand how the body moves over time, let's calculate the displacement [tex]\(x\)[/tex] for different values of [tex]\(t\)[/tex]. Here are the results for [tex]\(t\)[/tex] values from 1 to 10:
- For [tex]\(t = 1\)[/tex]:
[tex]\[ x = (1 - 3)^2 = (-2)^2 = 4 \][/tex]
- For [tex]\(t = 2\)[/tex]:
[tex]\[ x = (2 - 3)^2 = (-1)^2 = 1 \][/tex]
- For [tex]\(t = 3\)[/tex]:
[tex]\[ x = (3 - 3)^2 = 0^2 = 0 \][/tex]
- For [tex]\(t = 4\)[/tex]:
[tex]\[ x = (4 - 3)^2 = 1^2 = 1 \][/tex]
- For [tex]\(t = 5\)[/tex]:
[tex]\[ x = (5 - 3)^2 = 2^2 = 4 \][/tex]
- For [tex]\(t = 6\)[/tex]:
[tex]\[ x = (6 - 3)^2 = 3^2 = 9 \][/tex]
- For [tex]\(t = 7\)[/tex]:
[tex]\[ x = (7 - 3)^2 = 4^2 = 16 \][/tex]
- For [tex]\(t = 8\)[/tex]:
[tex]\[ x = (8 - 3)^2 = 5^2 = 25 \][/tex]
- For [tex]\(t = 9\)[/tex]:
[tex]\[ x = (9 - 3)^2 = 6^2 = 36 \][/tex]
- For [tex]\(t = 10\)[/tex]:
[tex]\[ x = (10 - 3)^2 = 7^2 = 49 \][/tex]
### Summary
Here are the displacements [tex]\(x\)[/tex] for each time [tex]\(t\)[/tex]:
- [tex]\(t = 1 \Rightarrow x = 4\)[/tex]
- [tex]\(t = 2 \Rightarrow x = 1\)[/tex]
- [tex]\(t = 3 \Rightarrow x = 0\)[/tex]
- [tex]\(t = 4 \Rightarrow x = 1\)[/tex]
- [tex]\(t = 5 \Rightarrow x = 4\)[/tex]
- [tex]\(t = 6 \Rightarrow x = 9\)[/tex]
- [tex]\(t = 7 \Rightarrow x = 16\)[/tex]
- [tex]\(t = 8 \Rightarrow x = 25\)[/tex]
- [tex]\(t = 9 \Rightarrow x = 36\)[/tex]
- [tex]\(t = 10 \Rightarrow x = 49\)[/tex]
This calculation shows how the displacement [tex]\(x\)[/tex] varies over time [tex]\(t\)[/tex]. The body starts with a displacement of 4 units at [tex]\(t = 1\)[/tex], decreases to 0 at [tex]\(t = 3\)[/tex], and then increases again as time progresses.
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