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Rewrite the formula for displacement, [tex]\( d = v_0 t + \frac{1}{2} a t^2 \)[/tex], to find [tex]\( a \)[/tex].

Arrange the equations in the correct sequence.

1. [tex]\( d - v_0 t = \frac{1}{2} a t^2 \)[/tex]
2. [tex]\( 2(d - v_0 t) = a t^2 \)[/tex]
3. [tex]\( a = \frac{2(d - v_0 t)}{t^2} \)[/tex]

Available tiles:
- [tex]\( d + v_0 t = \frac{1}{2} a t^2 \)[/tex]
- [tex]\( 2(d + v_0 t) = a t^2 \)[/tex]
- [tex]\( 2(d - v_0 t) = a t^2 \)[/tex]
- [tex]\( d - v_0 t = \frac{1}{2} a t^2 \)[/tex]
- [tex]\( a = \frac{2(d + v_0 t)}{t^2} \)[/tex]
- [tex]\( a = \frac{2(d - v_0 t)}{t^2} \)[/tex]

Sagot :

GinnyAnswer
Sure! Let's arrange the equations in the correct sequence to isolate [tex]\( a \)[/tex] in the given formula [tex]\( d = v_0 t + \frac{1}{2} a t^2 \)[/tex].

Step 1: Subtract [tex]\( v_0 t \)[/tex] from both sides to isolate the term with [tex]\( a \)[/tex]:
[tex]\[ d - v_0 t = \frac{1}{2} a t^2 \][/tex]

Step 2: Multiply both sides by 2 to eliminate the fraction:
[tex]\[ 2 (d - v_0 t) = a t^2 \][/tex]

Step 3: Divide both sides by [tex]\( t^2 \)[/tex] to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{2 (d - v_0 t)}{t^2} \][/tex]

Thus, the correct sequence is:
1. [tex]\( d - v_0 t = \frac{1}{2} a t^2 \)[/tex]
2. [tex]\( 2 (d - v_0 t) = a t^2 \)[/tex]
3. [tex]\( a = \frac{2 (d - v_0 t)}{t^2} \)[/tex]
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