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distance (meters) first graph: time (seconds) on x - axis, distance (me…

Question

distance (meters) first graph: time (seconds) on x - axis, distance (meters) on y - axis, a line from (0,0) to (3, ~40) second graph: distance (meters) on x - axis, time (seconds) on y - axis, a line from (0,0) to (3, ~40) table: the time and distance of a falling object

time (s)distance (m)
14.9
219.6
344.1

Explanation:

Response

Since the problem is not clearly stated, I'll assume a common question related to these graphs and table, like determining which graph correctly represents the motion of the falling object (or analyzing the relationship between time and distance).

For the first graph (Distance vs Time, top one):
  • The slope (speed) is calculated as $\frac{\Delta \text{Distance}}{\Delta \text{Time}}$. For points (1,10) and (2,30), slope is $\frac{30 - 10}{2 - 1}=20$ m/s. But from the table, at t=1s, distance is 4.9m, t=2s is 19.6m, t=3s is 44.1m. The relationship here is $d = 4.9t^2$ (free - fall, $d=\frac{1}{2}gt^2$, $g = 9.8$ m/s², so $\frac{1}{2}\times9.8=4.9$), which is a quadratic relationship, not linear. So the first graph (linear Distance - Time) is incorrect as it shows constant speed (linear), but free - fall has constant acceleration (quadratic distance - time).
For the second graph (Time vs Distance, bottom one):
  • Let's check the slope. If we take points (1,10) and (2,30), slope is $\frac{30 - 10}{2 - 1}=20$ s/m (time per distance), which is also a linear relationship, but the table shows a quadratic relationship between time and distance.
Analyzing the table:

The formula for the distance of a freely falling object is $d=\frac{1}{2}gt^2$, where $g = 9.8$ m/s². So $d = 4.9t^2$. Let's verify:

  • For $t = 1$: $d=4.9\times1^2 = 4.9$ m (matches the table).
  • For $t = 2$: $d=4.9\times2^2=4.9\times4 = 19.6$ m (matches the table).
  • For $t = 3$: $d=4.9\times3^2=4.9\times9 = 44.1$ m (matches the table).

If the question was about which graph is correct, neither of the two given graphs (which are linear) correctly represent the motion of the falling object, because the distance - time relationship for free - fall is quadratic ($d\propto t^2$), not linear. The table shows the correct (quadratic) relationship.

If you have a specific question (like finding speed, acceleration, or identifying the correct graph), please clarify.

Answer:

Since the problem is not clearly stated, I'll assume a common question related to these graphs and table, like determining which graph correctly represents the motion of the falling object (or analyzing the relationship between time and distance).

For the first graph (Distance vs Time, top one):
  • The slope (speed) is calculated as $\frac{\Delta \text{Distance}}{\Delta \text{Time}}$. For points (1,10) and (2,30), slope is $\frac{30 - 10}{2 - 1}=20$ m/s. But from the table, at t=1s, distance is 4.9m, t=2s is 19.6m, t=3s is 44.1m. The relationship here is $d = 4.9t^2$ (free - fall, $d=\frac{1}{2}gt^2$, $g = 9.8$ m/s², so $\frac{1}{2}\times9.8=4.9$), which is a quadratic relationship, not linear. So the first graph (linear Distance - Time) is incorrect as it shows constant speed (linear), but free - fall has constant acceleration (quadratic distance - time).
For the second graph (Time vs Distance, bottom one):
  • Let's check the slope. If we take points (1,10) and (2,30), slope is $\frac{30 - 10}{2 - 1}=20$ s/m (time per distance), which is also a linear relationship, but the table shows a quadratic relationship between time and distance.
Analyzing the table:

The formula for the distance of a freely falling object is $d=\frac{1}{2}gt^2$, where $g = 9.8$ m/s². So $d = 4.9t^2$. Let's verify:

  • For $t = 1$: $d=4.9\times1^2 = 4.9$ m (matches the table).
  • For $t = 2$: $d=4.9\times2^2=4.9\times4 = 19.6$ m (matches the table).
  • For $t = 3$: $d=4.9\times3^2=4.9\times9 = 44.1$ m (matches the table).

If the question was about which graph is correct, neither of the two given graphs (which are linear) correctly represent the motion of the falling object, because the distance - time relationship for free - fall is quadratic ($d\propto t^2$), not linear. The table shows the correct (quadratic) relationship.

If you have a specific question (like finding speed, acceleration, or identifying the correct graph), please clarify.