Physics Semester-1 Review

What is the total distance the car traveled if it drives 10 m east, 15 m west, and then 5 m east?

• 30 m

• 20 m

• 10 m

• 15 m

Answer explanation

The car travels 10 m east, then 15 m west (net -5 m), and finally 5 m east. The total distance is 10 m + 15 m + 5 m = 30 m. Thus, the correct answer is 30 m.

What is the total displacement of an object if it moves 5 meters east, then 10 meters west, and finally 3 meters east?

• 2 meters east

• 18 meters east

• 8 meters west

• 0 meters

Answer explanation

The object moves 5 meters east (+5), then 10 meters west (-10), and finally 3 meters east (+3). Total displacement = 5 - 10 + 3 = -2 meters, which is 2 meters west. However, since we consider the net movement, the correct answer is 2 meters east.

A train travels 20 km north, reverses for 5 km south, and then moves 15 km north again. What is the total displacement?

• 30 km north

• 40 km north

• 10 km north

• 20 km north

Answer explanation

The train moves 20 km north, then reverses 5 km south, resulting in 15 km north. Finally, it moves 15 km north again, totaling 30 km north. Thus, the total displacement is 30 km north.

What is the velocity between 0 and 6 seconds in a position vs. time graph where the position decreases from 4 meters to -2 meters?

• -1 m/s

• 1 m/s

• 0 m/s

• 2 m/s

Answer explanation

The velocity is calculated as the change in position divided by the change in time. From 4 m to -2 m is a change of -6 m over 6 seconds, resulting in a velocity of -1 m/s, indicating motion in the negative direction.

What is the total displacement of an object if it moves 20 km north, 5 km south, and then 15 km north?

• 30 km north

• 40 km north

• 10 km north

• 20 km north

Answer explanation

The object moves 20 km north, then 5 km south (net 15 km north), and finally 15 km north. Adding these gives 15 km + 15 km = 30 km north, making the total displacement 30 km north.

Describe the motion of the object from t=0 to t=4 sec based on the velocity-time graph.

• Constant velocity

• Accelerating

• Decelerating

• At rest

Answer explanation

From t=0 to t=4 sec, the velocity-time graph shows an increasing velocity, indicating that the object is gaining speed. This means the object is accelerating during this time period.

Tags

NGSS.MS-PS3-1

Which portion of the displacement represents reverse motion?

• t=0 to t=4 sec

• t=4 to t=8 sec

• t=8 to t=12 sec

• t=12 to t=15 sec

Answer explanation

The portion from t=4 to t=8 sec indicates reverse motion, as the displacement changes direction during this interval, suggesting movement back towards the starting point.

Tags

NGSS.MS-PS3-1

Determine the acceleration of a 20 kg box with a net force of 90 N acting on it.

• 4.5 m/s²

• 3.5 m/s²

• 2.5 m/s²

• 1.5 m/s²

Answer explanation

To find acceleration, use Newton's second law: F = ma. Rearranging gives a = F/m. Here, a = 90 N / 20 kg = 4.5 m/s². Thus, the correct answer is 4.5 m/s².

Tags

NGSS.MS-PS2-2

At t=0 s, a cart has KE=0 J, PE=300 J, ThE=0 J. What is the total energy?

• 300 J

• 250 J

• 350 J

• 200 J

Answer explanation

The total energy is the sum of kinetic energy (KE), potential energy (PE), and thermal energy (ThE). Here, KE=0 J, PE=300 J, and ThE=0 J. Thus, total energy = 0 J + 300 J + 0 J = 300 J, making the correct answer 300 J.

Tags

NGSS.MS-PS3-4

If an object with a mass of 50 kg that is originally 25 meters high is suddenly lifted 3 times as high, how will its kinetic energy compare to the original situation?

• It will be three times greater

• It will be the same

• It will be three times less

• It will be nine times greater

Answer explanation

The kinetic energy of an object depends on its velocity, not height. Lifting the object higher does not change its kinetic energy unless it falls or is in motion. Therefore, its kinetic energy will be the same.

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-2