Which of the equations below are dimensionally correct? (where v is velocity, x is

Question

position, a is acceleration, and t is time)𝑥
=
𝑥
𝑖
+
𝑣
𝑥
𝑖
𝑡
+
1
2
𝑎
𝑡
2
x=x
i
​
+v
xi
​
t+
2
1
​
at
2

Answer 1

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.

Let’s 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.