Question

A toy car with charge $q$ moves on a frictionless horizontal plane surface under the influence of a uniform electric field $E$ . Due to the force $q\,\vec{E}$ , its velocity increases from $0 \,to\, 6 m/s^{-1}$ 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. $1.5\, ms^{-1} , 3 ms^{-1}$
• B. $2 ms^{-1} , 4 ms^{-1}$
• C. $1 ms^{-1} , 3.5 ms^{-1}$
• D. $1 ms^{-1} , 3 ms^{-1}$

Answer 1

The Correct Option is D

To find the average velocity and average speed of the toy car between \(0\) to \(3\) seconds, we must analyze its motion step-by-step:

1. **Initial Motion (0 to 1 second):**
   The toy car starts from rest (\(u = 0\)) and accelerates under the force \(q\vec{E}\). Its velocity increases from \(0\) to \(6 \, \text{m/s}\) in one second. Using the formula for acceleration: a = \frac{\Delta v}{\Delta t} = \frac{6 - 0}{1} = 6 \, \text{m/s}^2 The displacement over this period can be calculated using: s_1 = ut + \frac{1}{2}at^2 = 0 \times 1 + \frac{1}{2} \times 6 \times 1^2 = 3 \, \text{m}
2. **Motion After Field Reversal (1 to 3 seconds):**
   At \(t = 1\) second, the field direction is reversed, so the acceleration is now \(-6 \, \text{m/s}^2\). The velocity at \(t = 1\) second (just before reversal) is \(6 \, \text{m/s}\). Calculate the velocity at \(t = 3\) seconds (after two more seconds) using: v = u + at = 6 + (-6) \times 2 = -6 \, \text{m/s} Calculate the total displacement from \(t = 1\) to \(3\) seconds: s_2 = vt + \frac{1}{2}at^2 = 6 \times 2 + \frac{1}{2} \times (-6) \times 2^2 = 12 - 12 = 0 \, \text{m}

Total displacement from \(0\) to \(3\) seconds:

s_{\text{total}} = s_1 + s_2 = 3 + 0 = 3 \, \text{m}

**Average Velocity:**
v_{\text{avg}} = \frac{s_{\text{total}}}{t_{\text{total}}} = \frac{3}{3} = 1 \, \text{m/s}

**Average Speed:**
Calculate the total path length which is the sum of magnitudes:
From \(0\) to \(1\) second: 3 \, \text{m}
From \(1\) to \(3\) seconds: The distance moved is still \(12 \, \text{m}\) (6 forward and 6 backward)
Total distance: 3 + 6 + 6 = 15 \, \text{m}

Average speed = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{15}{3} = 3 \, \text{m/s}

Therefore, the average velocity and the average speed are 1 \, \text{m/s} and 3 \, \text{m/s} respectively. The correct answer is 1 \, \text{m/s}, 3 \, \text{m/s}.