To determine which equation can be used to solve for acceleration, we will analyze each equation given.
1. Equation: [tex]\( t = \frac{\Delta v}{a} \)[/tex]
Let's solve for [tex]\( a \)[/tex]:
[tex]\[
a = \frac{\Delta v}{t}
\][/tex]
This equation can indeed be rearranged to solve for [tex]\( a \)[/tex].
2. Equation: [tex]\( v = a t - v_i \)[/tex]
Let's solve for [tex]\( a \)[/tex]:
[tex]\[
a t = v + v_i \quad \text{(adding \( v_i \) to both sides)}
\][/tex]
[tex]\[
a = \frac{v + v_i}{t}
\][/tex]
This equation can also solve for [tex]\( a \)[/tex], but it requires knowledge of [tex]\( v \)[/tex] (final velocity) and [tex]\( v_i \)[/tex] (initial velocity).
3. Equation: [tex]\( a = \frac{d}{t} \)[/tex]
This equation is already solved for [tex]\( a \)[/tex], indicating that the acceleration is the distance ([tex]\( d \)[/tex]) divided by the time ([tex]\( t \)[/tex]). However, this form suggests another context, such as uniform motion, and [tex]\( d \)[/tex] represents displacement.
4. Equation: [tex]\( \Delta v = a t \)[/tex]
Let's solve for [tex]\( a \)[/tex]:
[tex]\[
a = \frac{\Delta v}{t}
\][/tex]
This equation can also be rearranged to solve for [tex]\( a \)[/tex].
Given these analyses:
- The first equation solved for [tex]\( a \)[/tex] yields [tex]\( a = \frac{\Delta v}{t} \)[/tex].
- The second equation, when rearranged, also provides [tex]\( a \)[/tex].
- The third equation is already in the form to find [tex]\( a \)[/tex].
- The fourth equation, upon rearrangement, likewise yields [tex]\( a = \frac{\Delta v}{t} \)[/tex].
Considering the best option for solving for acceleration within typical kinematic contexts, the closest match from the equations provided gives us direct and clear use to find acceleration, thus pointing to:
[tex]\[
\Delta v = \frac{a}{t}
\][/tex]
This evaluates closely to solving for [tex]\( a \)[/tex] directly in kinematic contexts:
Hence, the correct choice is:
[tex]\[
4 \quad (\Delta v = a t)
\][/tex]
