. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of (a) 0, (b) 0, (c) 0, (d) 0, (e) 0, and (f) c?

Answer:

See Explanation

Explanation:

Given

[tex]s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}[/tex]

Solving (a): Units and dimension of [tex]s_0[/tex]

From the question, we understand that:

[tex]s \to L[/tex] --- length

[tex]t \to T[/tex] --- time

Remove the other terms of the equation, we have:

[tex]s=s_0[/tex]

Rewrite as:

[tex]s_0=s[/tex]

This implies that [tex]s_0[/tex] has the same unit and dimension as [tex]s[/tex]

Hence:

[tex]s_0 \to L[/tex] --- dimension

[tex]s_o \to[/tex] Length (meters, kilometers, etc.)

Solving (b): Units and dimension of [tex]v_0[/tex]

Remove the other terms of the equation, we have:

[tex]s=v_0t[/tex]

Rewrite as:

[tex]v_0t = s[/tex]

Make [tex]v_0[/tex] the subject

[tex]v_0 = \frac{s}{t}[/tex]

Replace s and t with their units

[tex]v_0 = \frac{L}{T}[/tex]

[tex]v_0 = LT^{-1}[/tex]

Hence:

[tex]v_0 \to LT^{-1}[/tex] --- dimension

[tex]v_0 \to[/tex] [tex]m/s[/tex] --- unit

Solving (c): Units and dimension of [tex]a_0[/tex]

Remove the other terms of the equation, we have:

[tex]s=\frac{a_0t^2}{2}[/tex]

Rewrite as:

[tex]\frac{a_0t^2}{2} = s_0[/tex]

Make [tex]a_0[/tex] the subject

[tex]a_0 = \frac{2s_0}{t^2}[/tex]

Replace s and t with their units [ignore all constants]

[tex]a_0 = \frac{L}{T^2}\\[/tex]

[tex]a_0 = LT^{-2[/tex]

Hence:

[tex]a_0 = LT^{-2[/tex] --- dimension

[tex]a_0 \to[/tex] [tex]m/s^2[/tex] --- acceleration

Solving (d): Units and dimension of [tex]j_0[/tex]

Remove the other terms of the equation, we have:

[tex]s=\frac{j_0t^3}{6}[/tex]

Rewrite as:

[tex]\frac{j_0t^3}{6} = s[/tex]

Make [tex]j_0[/tex] the subject

[tex]j_0 = \frac{6s}{t^3}[/tex]

Replace s and t with their units [Ignore all constants]

[tex]j_0 = \frac{L}{T^3}[/tex]

[tex]j_0 = LT^{-3}[/tex]

Hence:

[tex]j_0 = LT^{-3}[/tex] --- dimension

[tex]j_0 \to[/tex] [tex]m/s^3[/tex] --- unit

Solving (e): Units and dimension of [tex]s_0[/tex]

Remove the other terms of the equation, we have:

[tex]s=\frac{S_0t^4}{24}[/tex]

Rewrite as:

[tex]\frac{S_0t^4}{24} = s[/tex]

Make [tex]S_0[/tex] the subject

[tex]S_0 = \frac{24s}{t^4}[/tex]

Replace s and t with their units [ignore all constants]

[tex]S_0 = \frac{L}{T^4}[/tex]

[tex]S_0 = LT^{-4[/tex]

Hence:

[tex]S_0 = LT^{-4[/tex] --- dimension

[tex]S_0 \to[/tex] [tex]m/s^4[/tex] --- unit

Solving (e): Units and dimension of [tex]c[/tex]

Ignore other terms of the equation, we have:

[tex]s=\frac{ct^5}{120}[/tex]

Rewrite as:

[tex]\frac{ct^5}{120} = s[/tex]

Make [tex]c[/tex] the subject

[tex]c = \frac{120s}{t^5}[/tex]

Replace s and t with their units [Ignore all constants]

[tex]c = \frac{L}{T^5}[/tex]

[tex]c = LT^{-5}[/tex]

Hence:

[tex]c \to LT^{-5}[/tex] --- dimension

[tex]c \to m/s^5[/tex] --- units

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