Question

three trucks start at rest, side-by-side. the trucks carry different mass loads. the three drivers press their accelerators at the same time, and the trucks experience equal strength net forces. the positions of the three trucks before starting and a short time later are modeled below. image of three trucks (g, r, b) at start and later positions rank the mass of each truck (including its load) from greatest to least. greatest mass r g b least mass

Explanation:

Step1: Recall Newton's Second Law

Newton's second law states \( F = ma \), where \( F \) is net force, \( m \) is mass, and \( a \) is acceleration. Given \( F \) is equal for all trucks, \( m \) and \( a \) are inversely proportional (\( m=\frac{F}{a} \); smaller \( a \) means larger \( m \)).

Step2: Analyze Acceleration from Displacement

All trucks start at rest (\( v_0 = 0 \)) and move for the same time \( t \). The displacement formula for constant acceleration is \( x = v_0t+\frac{1}{2}at^2 \), so \( x=\frac{1}{2}at^2 \) (since \( v_0 = 0 \)). Rearranging, \( a=\frac{2x}{t^2} \). Displacement \( x \) is proportional to \( a \) (same \( t \)).

From the diagram:
- Red truck (R) has the largest displacement.
- Green truck (G) has the smallest displacement.
- Blue truck (B) has a displacement between R and G.

Step3: Relate Displacement to Mass

Since \( x \propto a \) (same \( t \)), and \( m \propto \frac{1}{a} \) (same \( F \)):
- Largest displacement (R) → largest \( a \) → smallest \( m \).
- Smallest displacement (G) → smallest \( a \) → largest \( m \).
- Displacement of B is between R and G → \( a \) of B is between R and G → \( m \) of B is between G and R.

Step4: Rank Masses

From largest to smallest mass: G (largest \( m \)) > B > R (smallest \( m \)). Wait, correction: Wait, no—wait, \( x=\frac{1}{2}at^2 \), so larger \( x \) means larger \( a \). Since \( m = F/a \), larger \( a \) means smaller \( m \). So:
- G has smallest \( x \) → smallest \( a \) → largest \( m \).
- B has middle \( x \) → middle \( a \) → middle \( m \).
- R has largest \( x \) → largest \( a \) → smallest \( m \).

So mass order: G (greatest) > B > R (least)? Wait, no—wait the diagram: Let me re-examine the diagram. Wait, the green truck (G) moves the least distance, red (R) moves the most, blue (B) is in between. So:

- G: least displacement → least acceleration (since \( x \propto a \) for same \( t \), \( v_0=0 \)) → since \( F=ma \), same \( F \), least \( a \) means greatest \( m \).
- R: most displacement → most acceleration → least \( m \).
- B: middle displacement → middle acceleration → middle \( m \).

So mass ranking: G (greatest) > B > R (least)? Wait, no—wait the problem's diagram: Wait, the green truck (G) is the top one, red (R) middle, blue (B) bottom. Wait, the green truck's displacement is the smallest (moved the least), red the most, blue in between. So:

Acceleration: R (most x) → highest a; G (least x) → lowest a; B (middle x) → middle a.

Mass: since \( m = F/a \), same F, so:

- G: lowest a → highest m.
- B: middle a → middle m.
- R: highest a → lowest m.

So the order from greatest to least mass is G, B, R? Wait, no—wait the initial thought was reversed. Let's confirm:

If F is same, then m = F/a. So smaller a (less acceleration) means larger m. Acceleration is determined by how much they moved (displacement) in same time. Since they start from rest, the one that moved the least (G) has the smallest acceleration (because x = 0.5*a*t², so smaller x means smaller a, same t). So G has smallest a → largest m. R moved the most → largest a → smallest m. B is in between. So mass order: G (greatest) > B > R (least). Wait, but let's check again.

Wait, the diagram: Green truck (G) is at the top lane. At start, all are at the dashed line. After time, G is just a bit ahead, R is far ahead, B is in the middle. So displacement: R > B > G. So x_R > x_B > x_G.

Since x = 0.5*a*t², same t, so a_R > a_B > a_G (because x is proportional to a).

Then m = F/a, same F, so m_G > m_B > m_R (because a is smallest for G, so m is largest; a is largest for R, so m is smallest).

So the order from greatest to least mass is G, B, R? Wait, but the options are R, G, B. Wait, no—the problem's ranking is "greatest mass" to "least mass". So the greatest mass is G, then B, then R. Wait, but the user's diagram: Let me re-express the displacement:

- Green (G): moved the least (smallest x)
- Blue (B): moved more than G, less than R
- Red (R): moved the most (largest x)

Thus, acceleration: a_R > a_B > a_G (since x ∝ a for same t, v0=0)

Mass: m_G > m_B > m_R (since m = F/a, same F)

So the ranking from greatest to least mass is G, B, R. Wait, but the problem's options are R, G, B as draggable items. Wait, maybe I misread the diagram. Wait, maybe the red truck is the one that moved the most, so it has the highest acceleration, hence lowest mass. Green moved the least, lowest acceleration, highest mass. Blue in between. So the order from greatest to least mass is G (greatest), B, R (least).

Answer:

greatest mass: G
middle: B
least mass: R

So the ranking from greatest to least mass is G, B, R.