Question
Which of the equations below are dimensionally correct? (where v is velocity, x is
position, a is acceleration, and t is time)π₯
=
π₯
π
+
π£
π₯
π
π‘
+
1
2
π
π‘
2
x=x
i
β
+v
xi
β
t+
2
1
β
at
2
position, a is acceleration, and t is time)π₯
=
π₯
π
+
π£
π₯
π
π‘
+
1
2
π
π‘
2
x=x
i
β
+v
xi
β
t+
2
1
β
at
2
Answers
GPT-4o mini
To determine if the equation
\[
x = x_i + v_{xi} t + \frac{1}{2} a t^2
\]
is dimensionally correct, we need to analyze the dimensions of each term in the equation.
Letβs define the dimensions of the variables involved:
- \(x\) (position) has the dimension of length: \([L]\)
- \(x_i\) (initial position) also has the dimension of length: \([L]\)
- \(v_{xi}\) (initial velocity) has the dimension of length per time: \([L][T^{-1}]\)
- \(t\) (time) has the dimension of time: \([T]\)
- \(a\) (acceleration) has the dimension of length per time squared: \([L][T^{-2}]\)
Now, we evaluate the dimensions of each term:
1. The left side \(x\) has the dimension:
\[
[L]
\]
2. The term \(x_i\) also has the dimension:
\[
[L]
\]
3. The term \(v_{xi} t\) has dimensions:
\[
[L][T^{-1}] \cdot [T] = [L]
\]
4. The term \(\frac{1}{2} a t^2\) has dimensions:
\[
\frac{1}{2} \cdot [L][T^{-2}] \cdot [T^2] = [L]
\]
Now, putting it all together:
The right side of the equation is:
\[
x_i + v_{xi} t + \frac{1}{2} a t^2
\]
This can be expressed as:
\[
[L] + [L] + [L] = [L]
\]
Since both sides of the equation have the dimension of length \([L]\), we can conclude that the equation is dimensionally correct.
**Final conclusion: The equation \(x = x_i + v_{xi} t + \frac{1}{2} a t^2\) is dimensionally correct.**
\[
x = x_i + v_{xi} t + \frac{1}{2} a t^2
\]
is dimensionally correct, we need to analyze the dimensions of each term in the equation.
Letβs define the dimensions of the variables involved:
- \(x\) (position) has the dimension of length: \([L]\)
- \(x_i\) (initial position) also has the dimension of length: \([L]\)
- \(v_{xi}\) (initial velocity) has the dimension of length per time: \([L][T^{-1}]\)
- \(t\) (time) has the dimension of time: \([T]\)
- \(a\) (acceleration) has the dimension of length per time squared: \([L][T^{-2}]\)
Now, we evaluate the dimensions of each term:
1. The left side \(x\) has the dimension:
\[
[L]
\]
2. The term \(x_i\) also has the dimension:
\[
[L]
\]
3. The term \(v_{xi} t\) has dimensions:
\[
[L][T^{-1}] \cdot [T] = [L]
\]
4. The term \(\frac{1}{2} a t^2\) has dimensions:
\[
\frac{1}{2} \cdot [L][T^{-2}] \cdot [T^2] = [L]
\]
Now, putting it all together:
The right side of the equation is:
\[
x_i + v_{xi} t + \frac{1}{2} a t^2
\]
This can be expressed as:
\[
[L] + [L] + [L] = [L]
\]
Since both sides of the equation have the dimension of length \([L]\), we can conclude that the equation is dimensionally correct.
**Final conclusion: The equation \(x = x_i + v_{xi} t + \frac{1}{2} a t^2\) is dimensionally correct.**