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
