Review - Modeling Motion Unit 1

Choose the time interval with constant positive acceleration

• 1 to 4 seconds

• 7 to 9 seconds

• 1 to 10 seconds

• 4 to 5 seconds

Answer explanation

The interval from 1 to 4 seconds shows a consistent increase in velocity, indicating constant positive acceleration. Other intervals either show no acceleration or negative acceleration.

Tags

NGSS.HS-PS2-1

What is the acceleration between t=9 s and t=10 s?

• (5.00 m/s-1.00 m/s)/(10s-9s) = (4 m/s)/(1s) = 4 m/s^2

• (5.00 m/s-1.00 m/s)/(10s-9s) = (6 m/s)/(1s) = 6 m/s^2

• (5.00 m/s-1.00 m/s)/(10s-9s) = (8 m/s)/(1s) = 8 m/s^2

• (5.00 m/s-1.00 m/s)/(10s-9s) = (6 m/s)/(1s) = 5 m/s^2

Answer explanation

To find acceleration, use the formula (final velocity - initial velocity) / time interval. Here, (5.00 m/s - 1.00 m/s) / (10s - 9s) = 4 m/s / 1s = 4 m/s², making the correct choice the first option.

Tags

NGSS.HS-PS2-1

What is the displacement between t=6 s and t=7 s?

• (11.0 m-9.0 m) = 1.00 m

• (13.0 m-12.0 m) = 1.00 m

• (12.0 m-9.0 m) = 1.00 m

• (9.0 m-7.0 m) = 2.00 m

Answer explanation

To find the displacement between t=6 s and t=7 s, we calculate the difference in position: (13.0 m - 12.0 m) = 1.00 m. This confirms that the correct answer is (13.0 m-12.0 m) = 1.00 m.

Choose the time interval with constant positive acceleration.

• 1 to 4 seconds

• 7 to 9 seconds

• 1 to 10 seconds

• 4 to 5 seconds

Answer explanation

The interval from 1 to 4 seconds shows a consistent increase in velocity, indicating constant positive acceleration. Other intervals either show no acceleration or negative acceleration.

Tags

NGSS.HS-PS2-1

Based on the data, which of the following options identifies what the plot of each graph represents, in order from Graph A to Graph C?

• A. Velocity vs. Time; Acceleration vs. Time; Distance vs. Time

• B. Acceleration vs. Time; Velocity vs. Time; Distance vs. Time

• C. Distance vs. Time; Velocity vs. Time; Acceleration vs. Time

• D. Distance vs. Time; Acceleration vs. Time; Velocity vs. Time

Answer explanation

The correct choice is C. Graph A represents Distance vs. Time, showing how distance changes over time. Graph B shows Velocity vs. Time, indicating speed changes. Graph C depicts Acceleration vs. Time, reflecting changes in velocity.

On an unspecified planet, the acceleration due to gravity is −0.4 m/s². How does this compare to the acceleration due to gravity on Earth? If two identical objects are dropped from the same height on both this planet and Earth, which will reach the ground in the shortest amount of time?

• The object on Earth will reach the ground first because Earth's gravity is stronger.

• The object on the unspecified planet will reach the ground first because its gravity is stronger.

• Both objects will reach the ground at the same time because gravity is the same.

• Neither object will reach the ground because gravity is zero.

Answer explanation

The acceleration due to gravity on Earth is approximately 9.81 m/s², which is much stronger than the −0.4 m/s² on the unspecified planet. Therefore, the object on Earth will reach the ground first.

Tags

NGSS.HS-PS2-4

Order each action from the least to greatest displacement.

• 3 < 4 < 1 < 5 < 2

• 3 < 5 < 1 < 4 < 2

• 4 < 3 < 2 < 5 < 1

• 2 < 4 < 1 < 5 < 3

Answer explanation

The correct order of displacement is based on the distance each action covers. The sequence 3 < 5 < 1 < 4 < 2 indicates that action 3 has the least displacement, followed by 5, then 1, 4, and finally 2 with the greatest displacement.

A soccer ball is slowly being kicked across a flat field. Based on the data provided, what is the average acceleration? Assume that at t = 0 minutes, the soccer ball is at rest.

• A. 0.0234 m/s²

• B. 0.00263 m/s²

• C. 0.0187 m/s²

• D. 0.00210 m/s²

Answer explanation

To find the average acceleration, use the formula a = (final velocity - initial velocity) / time. Given the ball starts from rest, the correct calculation yields an average acceleration= change of velocity 0.562-0/300 so final acceleration is 0.0187 m/s², matching choice C.

Tags

NGSS.HS-PS2-1

If an object’s velocity increases from 13.2 m/s to 15.1 m/s in 8.00 seconds, what is the acceleration of the object?

• 1.89 m/s²

• -0.250 m/s²

• -1.65 m/s²

• 0.238 m/s²

Answer explanation

To find acceleration, use the formula: a = (final velocity - initial velocity) / time. Here, a = (15.1 m/s - 13.2 m/s) / 8.00 s = 0.238 m/s². Thus, the correct answer is 0.238 m/s².

Tags

NGSS.HS-PS2-1

1. Choose the option that identifies the correct terms, in order of their appearance in the sentences.

The _____ velocity is the _____ of the tangent line on a _____ vs. time graph.

• average; slope; acceleration

• instantaneous; integral; position

• average; integral; position

• instantaneous; slope; position

Answer explanation

The correct choice is 'instantaneous; slope; position' because instantaneous velocity refers to the velocity at a specific moment, which is represented by the slope of the tangent line on a position vs. time graph.