Question

drag each tile to the correct box. not all tiles will be used
arrange the equations in the correct sequence to rewrite the formula for displacement, $d = vt - \frac{1}{2}at^2$, to find $a$. in the formula, $d$ is displacement, $v$ is final velocity, $a$ is acceleration, and $t$ is time.
$2\left(vt - d\
ight) = at^2$
$a = \frac{2(d - vt)}{t^2}$
$2\left(d - vt\
ight) = at^2$
$vt - d = \frac{1}{2}at^2$
$d - vt = \frac{1}{2}at^2$
$a = \frac{2(vt - d)}{t^2}$

Explanation:

Step1: Start with the original formula

We have the displacement formula \( d = vt - \frac{1}{2}at^2 \). First, we want to isolate the term with \( a \). Subtract \( vt \) from both sides:
\( d - vt = -\frac{1}{2}at^2 \), or multiplying both sides by -1 (which is equivalent to rearranging terms): \( vt - d = \frac{1}{2}at^2 \). Wait, actually, let's do it step by step. Let's move \( vt \) to the left: \( d - vt = -\frac{1}{2}at^2 \), but maybe a better first step is to subtract \( d \) from both sides and add \( \frac{1}{2}at^2 \) to both sides? Wait, no, let's follow the tiles. The first step should be manipulating the original equation. Let's start with \( d = vt - \frac{1}{2}at^2 \). Subtract \( vt \) from both sides: \( d - vt = -\frac{1}{2}at^2 \). Multiply both sides by -1: \( vt - d = \frac{1}{2}at^2 \). Wait, but one of the tiles is \( d - vt = \frac{1}{2}at^2 \) (with a negative sign). Wait, maybe I made a sign error. Let's do it again:

Original formula: \( d = vt - \frac{1}{2}at^2 \)

Subtract \( vt \) from both sides: \( d - vt = -\frac{1}{2}at^2 \)

Multiply both sides by -1: \( vt - d = \frac{1}{2}at^2 \). So the first tile in the sequence should be \( vt - d = \frac{1}{2}at^2 \)? Wait, no, let's check the tiles. Wait, the tiles include \( d - vt = \frac{1}{2}at^2 \) (which is \( d - vt = \frac{1}{2}at^2 \), which is equivalent to \( - (vt - d) = \frac{1}{2}at^2 \), or \( vt - d = -\frac{1}{2}at^2 \), but maybe the first step is \( vt - d = \frac{1}{2}at^2 \)? Wait, no, let's proceed.

Wait, the correct first step: Let's solve for \( a \). Start with \( d = vt - \frac{1}{2}at^2 \).

Step 1: Subtract \( vt \) from both sides: \( d - vt = -\frac{1}{2}at^2 \).

Step 2: Multiply both sides by -2: \( -2(d - vt) = at^2 \), which is \( 2(vt - d) = at^2 \) (since \( -2(d - vt) = 2(vt - d) \)). Wait, but let's check the tiles. The tiles have \( vt - d = \frac{1}{2}at^2 \), \( d - vt = \frac{1}{2}at^2 \), etc. Wait, maybe the first step is \( vt - d = \frac{1}{2}at^2 \)? No, let's do it step by step with the tiles.

Wait, the first tile in the sequence (top box) should be the first manipulation. Let's list the tiles:

1. \( 2(vt - d) = at^2 \)
2. \( a = \frac{2(d - vt)}{t^2} \)
3. \( 2(d - vt) = at^2 \)
4. \( vt - d = \frac{1}{2}at^2 \)
5. \( d - vt = \frac{1}{2}at^2 \)
6. \( a = \frac{2(vt - d)}{t^2} \)

Wait, let's start with the original formula \( d = vt - \frac{1}{2}at^2 \).

Step 1: Subtract \( vt \) from both sides: \( d - vt = -\frac{1}{2}at^2 \). Multiply both sides by -1: \( vt - d = \frac{1}{2}at^2 \). So the first tile is \( vt - d = \frac{1}{2}at^2 \).

Step 2: Multiply both sides by 2 to eliminate the fraction: \( 2(vt - d) = at^2 \). So the second tile is \( 2(vt - d) = at^2 \). Wait, but there's a tile \( 2(d - vt) = at^2 \). Wait, no, \( 2(vt - d) = at^2 \) is correct here. Wait, but maybe I messed up the sign. Let's check:

From \( vt - d = \frac{1}{2}at^2 \), multiply both sides by 2: \( 2(vt - d) = at^2 \). That's one tile.

Then, to solve for \( a \), divide both sides by \( t^2 \): \( a = \frac{2(vt - d)}{t^2} \). But wait, there's a tile \( a = \frac{2(d - vt)}{t^2} \), which is equivalent to \( a = -\frac{2(vt - d)}{t^2} \), which would be wrong. Wait, no, let's go back.

Wait, original formula: \( d = vt - \frac{1}{2}at^2 \). Let's solve for \( a \):

1. Subtract \( vt \) from both sides: \( d - vt = -\frac{1}{2}at^2 \).
2. Multiply both sides by -2: \( -2(d - vt) = at^2 \), which is \( 2(vt - d) = at^2 \) (since \( -2(d - vt) = 2(vt - d) \)).
3. Now, divide both sides by \( t^2 \) (assuming \( t eq 0 \)): \( a = \frac{2(vt - d)}{t^2} \). But wait, the tile \( a = \frac{2(d - vt)}{t^2} \) is also present. Wait, maybe I made a sign error in the first step. Let's do it again:

Original formula: \( d = vt - \frac{1}{2}at^2 \)

Add \( \frac{1}{2}at^2 \) to both sides: \( d + \frac{1}{2}at^2 = vt \)

Subtract \( d \) from both sides: \( \frac{1}{2}at^2 = vt - d \)

Multiply both sides by 2: \( at^2 = 2(vt - d) \)

Then divide by \( t^2 \): \( a = \frac{2(vt - d)}{t^2} \). But the tile \( 2(d - vt) = at^2 \) would be \( at^2 = 2(d - vt) \), which is \( at^2 = -2(vt - d) \), so \( a = \frac{-2(vt - d)}{t^2} = \frac{2(d - vt)}{t^2} \). Ah, here's the confusion. So there are two possible paths, but let's follow the tiles.

Wait, the correct sequence should be:

1. Start with \( d = vt - \frac{1}{2}at^2 \). Subtract \( vt \) from both sides: \( d - vt = -\frac{1}{2}at^2 \). Multiply both sides by -1: \( vt - d = \frac{1}{2}at^2 \) (tile: \( vt - d = \frac{1}{2}at^2 \)).

2. Multiply both sides by 2: \( 2(vt - d) = at^2 \) (tile: \( 2(vt - d) = at^2 \)).

3. Then divide both sides by \( t^2 \): \( a = \frac{2(vt - d)}{t^2} \) (tile: \( a = \frac{2(vt - d)}{t^2} \)). But wait, the tile \( a = \frac{2(d - vt)}{t^2} \) is also there. Wait, maybe the first step is \( d - vt = \frac{1}{2}at^2 \) (with a negative sign). Let's check:

From \( d = vt - \frac{1}{2}at^2 \), subtract \( vt \): \( d - vt = -\frac{1}{2}at^2 \), which is \( d - vt = \frac{1}{2}(-at^2) \), but the tile \( d - vt = \frac{1}{2}at^2 \) would imply \( -at^2/2 = at^2/2 \), which is only true if \( a = 0 \), so that's wrong. Therefore, the correct first step is \( vt - d = \frac{1}{2}at^2 \), then multiply by 2: \( 2(vt - d) = at^2 \), then solve for \( a \): \( a = \frac{2(vt - d)}{t^2} \). But wait, the tile \( a = \frac{2(d - vt)}{t^2} \) is equivalent to \( a = -\frac{2(vt - d)}{t^2} \), which would be incorrect. Wait, maybe the problem has a sign error, or I'm misinterpreting the tiles.

Wait, let's look at the tiles again. The tiles are:

1. \( 2(vt - d) = at^2 \)
2. \( a = \frac{2(d - vt)}{t^2} \)
3. \( 2(d - vt) = at^2 \)
4. \( vt - d = \frac{1}{2}at^2 \)
5. \( d - vt = \frac{1}{2}at^2 \)
6. \( a = \frac{2(vt - d)}{t^2} \)

Let's follow the correct algebraic steps:

1. Start with \( d = vt - \frac{1}{2}at^2 \).

2. Subtract \( vt \) from both sides: \( d - vt = -\frac{1}{2}at^2 \). Multiply both sides by -1: \( vt - d = \frac{1}{2}at^2 \) (tile 4: \( vt - d = \frac{1}{2}at^2 \)).

3. Multiply both sides by 2: \( 2(vt - d) = at^2 \) (tile 1: \( 2(vt - d) = at^2 \)).

4. Divide both sides by \( t^2 \) (assuming \( t eq 0 \)): \( a = \frac{2(vt - d)}{t^2} \) (tile 6: \( a = \frac{2(vt - d)}{t^2} \)).

But wait, the tile \( a = \frac{2(d - vt)}{t^2} \) is also present. Let's check the sign: \( 2(d - vt) = 2(-(vt - d)) = -2(vt - d) \), so \( a = \frac{2(d - vt)}{t^2} = \frac{-2(vt - d)}{t^2} \), which would be equivalent to \( a = -\frac{2(vt - d)}{t^2} \), which contradicts the original formula. Therefore, the correct sequence should be:

First tile: \( vt - d = \frac{1}{2}at^2 \)

Second tile: \( 2(vt - d) = at^2 \)

Third tile: \( a = \frac{2(vt - d)}{t^2} \)

But wait, the problem shows a vertical sequence with three boxes. Let's confirm:

1. First box (top): Start with manipulating the original equation to get \( vt - d = \frac{1}{2}at^2 \) (tile 4).

2. Second box (middle): Multiply by 2 to get \( 2(vt - d) = at^2 \) (tile 1).

3. Third box (bottom): Divide by \( t^2 \) to get \( a = \frac{2(vt - d)}{t^2} \) (tile 6).

Wait, but there's a tile \( 2(d - vt) = at^2 \) (tile 3) and \( a = \frac{2(d - vt)}{t^2} \) (tile 2). Let's see if that's a possible path. Suppose we start with \( d = vt - \frac{1}{2}at^2 \), subtract \( vt \): \( d - vt = -\frac{1}{2}at^2 \), multiply by -2: \( -2(d - vt) = at^2 \), which is \( 2(vt - d) = at^2 \), not \( 2(d - vt) = at^2 \). So \( 2(d - vt) = at^2 \) would be \( at^2 = 2(d - vt) \), which is \( at^2 = -2(vt - d) \), so \( a = \frac{-2(vt - d)}{t^2} = \frac{2(d - vt)}{t^2} \). Let's check if that's correct. If we plug \( a = \frac{2(d - vt)}{t^2} \) into the original formula:

\( d = vt - \frac{1}{2} \cdot \frac{2(d - vt)}{t^2} \cdot t^2 = vt - (d - vt) = vt - d + vt = 2vt - d \), which is not equal to \( d \) unless \( 2vt - d = d \) → \( 2vt = 2d \) → \( vt = d \), which is not general. Therefore, that path is incorrect.

Therefore, the correct sequence is:

1. \( vt - d = \frac{1}{2}at^2 \) (tile 4)

2. \( 2(vt - d) = at^2 \) (tile 1)

3. \( a = \frac{2(vt - d)}{t^2} \) (tile 6)

Answer:

The correct sequence (from top to bottom) is:

1. \( vt - d = \frac{1}{2}at^2 \)
2. \( 2(vt - d) = at^2 \)
3. \( a = \frac{2(vt - d)}{t^2} \)

(Note: If the tiles are to be dragged, the order is: first tile \( vt - d = \frac{1}{2}at^2 \), then \( 2(vt - d) = at^2 \), then \( a = \frac{2(vt - d)}{t^2} \).)