A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field \(\vec{\text{E}}\). Due to the force q \(\vec{\text{E}}\), 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

Concept:

The three equations of motion are-

v = u +at

v² - u² = 2as

s = ut + (1/2) at²

where s, u, v, a, t are distance travelled, initial velocity, final velocity, acceleration and time taken respectively.

\(Average\ speed\ =\frac{Total\ distance\ }{Total\ time\ taken}\)

\(Average\ velocity\ =\frac{Total\ displacement\ }{Total\ time\ taken}\)

Calculation:

Acceleration \(a = \frac{{6 - 0}}{1} = 6m{s^{ - 2}}\)

For t = 0 to t = 1 s,

\({S_1} = \frac{1}{2} \times 6{\left( 1 \right)^2} = 3m\) ...(i)

For t = 1 s to t = 2 s, 

\({S_2} = 6.1 - \frac{1}{2} \times 6{\left( 1 \right)^2} = 3m\) ...(ii)

For t = 2 s to t = 3 s,

\({S_3} = 0 - \frac{1}{2} \times 6{\left( 1 \right)^2} = - 3m\) ...(iii)

Total displacement S = S₁ + S₂ + S₃ = 3 m

Average velocity \(= \frac{3}{3} = 1m{s^{ - 1}}\)

Total distance travelled = 9 m

Average speed \( = \frac{9}{3} = 3m{s^{ - 1}}\)