Question

question 12 the position, x, of an object is given by the equation x = a + bt+ct², where t refers to time. what are the dimensions of a, b, and c? show your work for credit.

Explanation:

Step1: Recall dimension - homogeneity principle

In an equation, all terms must have the same dimensions. The dimension of position \(x\) is length \(L\).

Step2: Analyze the dimension of term \(A\)

Since \(x = A+Bt + Ct^{2}\), and \(x\) has dimension \(L\), for the term \(A\) to be added to \(Bt\) and \(Ct^{2}\), \(A\) must have the dimension of length. So, \([A]=L\).

Step3: Analyze the dimension of term \(Bt\)

The dimension of \(t\) is time \(T\). Let the dimension of \(B\) be \([B]\). Since \(Bt\) must have the dimension of length (to be added to \(A\) and \(Ct^{2}\)), we have \([B][T]=L\). Solving for \([B]\), we get \([B]=\frac{L}{T}\).

Step4: Analyze the dimension of term \(Ct^{2}\)

Let the dimension of \(C\) be \([C]\). Since \(Ct^{2}\) must have the dimension of length, and the dimension of \(t^{2}\) is \(T^{2}\), we have \([C][T^{2}]=L\). Solving for \([C]\), we get \([C]=\frac{L}{T^{2}}\).

Answer:

\([A]=L\), \([B]=\frac{L}{T}\), \([C]=\frac{L}{T^{2}}\)