Question

1. -/3 points assume the equation x = at³+bt describes the motion of a particular object, with x having the dimension of length and t having the dimension of time. determine the dimensions of the constants a and b. (use the following as necessary: l and t, where l is the unit of length and t is the unit of time.) a = b = (b) determine the dimensions of the derivative dx/dt = 3at²+ b. (use the following as necessary: l and t, where l is the unit of length and t is the unit of time.) dx/dt =

Explanation:

Step1: Analyze the dimensions of \(x\)

Given \(x = At^{3}+Bt\), the dimensions of \(x\) is \(L\). For the right - hand side, the dimensions of each term must be \(L\). For the term \(At^{3}\), if the dimension of \(t\) is \(T\), then \([At^{3}]=L\). Let \([A]=L/T^{3}\). For the term \(Bt\), \([Bt]=L\), so \([B]=L/T\).

Step2: Differentiate \(x\) with respect to \(t\)

We have \(x = At^{3}+Bt\). Differentiating using the power rule \(\frac{d}{dt}(t^{n})=nt^{n - 1}\), we get \(\frac{dx}{dt}=3At^{2}+B\).

Step3: Determine the dimensions of \(\frac{dx}{dt}\)

The dimension of \(3At^{2}\): Substitute \([A]=L/T^{3}\) into \([3At^{2}]\), we have \([3At^{2}]=3\times\frac{L}{T^{3}}\times T^{2}=\frac{L}{T}\). The dimension of \(B\) is \(\frac{L}{T}\). So \([\frac{dx}{dt}]=\frac{L}{T}\).

Answer:

\([A]=\frac{L}{T^{3}}\), \([B]=\frac{L}{T}\), \([\frac{dx}{dt}]=\frac{L}{T}\)