Question

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field \(\overrightarrow{E}\) . Due to the force q\(\overrightarrow{E}\), its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

• A. 2m/s, 4m/s
• B. 1m/s, 3.5m/s
• C. 1m/s, 3m/s
• D. 1.5m/s, 3m/s

Answer 1

The Correct Option is C

To determine the average velocity and average speed of the toy car between 0 to 3 seconds, we need to evaluate the motion of the car in the context of the changes in the electric field direction.

1. Initially, the car moves with an acceleration caused by the electric field, which increases its velocity from 0 to 6 m/s in 1 second.

The acceleration \( a \) is given by the formula:

a = \frac{\Delta v}{\Delta t} = \frac{6 \, \text{m/s} - 0 \, \text{m/s}}{1 \, \text{s}} = 6 \, \text{m/s}^2

2. At the end of the first second, the electric field's direction is reversed. Hence, the car now decelerates under this reversed field.
3. The velocity from 1 to 2 seconds is decreased by the same acceleration of 6 m/s². Hence,

The velocity after the next second (2 seconds in total) is:

v = 6 \, \text{m/s} - (6 \, \text{m/s}^2 \times 1 \, \text{s}) = 0 \, \text{m/s}

4. The car continues to move under the same reversed field for another second (from 2 to 3 seconds), reaching a velocity of:

v = 0 \, \text{m/s} - (6 \, \text{m/s}^2 \times 1 \, \text{s}) = -6 \, \text{m/s}

5. Now, let's calculate the average velocity and average speed over the 3 seconds:

Average Velocity:

The displacement can be calculated as the sum of individual displacements in each phase:

• Displacement in the first second: d_1 = \frac{1}{2} \times 6 \, \text{m/s} \times 1 \, \text{s} = 3 \, \text{m}
• Displacement in the second second: d_2 = \frac{1}{2} \times (6 \, \text{m/s} + 0 \, \text{m/s}) \times 1 \, \text{s} = 3 \, \text{m}
• Displacement in the third second: d_3 = \frac{1}{2} \times (0 \, \text{m/s} + (-6) \, \text{m/s}) \times 1 \, \text{s} = -3 \, \text{m}

Total displacement = 3 \, \text{m} + 3 \, \text{m} - 3 \, \text{m} = 3 \, \text{m}

Average velocity = \frac{\text{Total displacement}}{\text{Total time}} = \frac{3 \, \text{m}}{3 \, \text{s}} = 1 \, \text{m/s}

Average Speed:

Total distance travelled = |d_1| + |d_2| + |d_3| = 3 \, \text{m} + 3 \, \text{m} + 3 \, \text{m} = 9 \, \text{m}

Average speed = \frac{\text{Total distance}}{\text{Total time}} = \frac{9 \, \text{m}}{3 \, \text{s}} = 3 \, \text{m/s}

6. Therefore, the average velocity is 1 m/s and the average speed is 3 m/s.

The correct choice is therefore: 1 m/s, 3 m/s.