Question

en
A toy car with charge q moves on a frictionless
horizontal plane surface under the influence of a
uniform electric field Ë. Due to the force g Ë , 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
[NEET-2018]
6.
(1) 2 m/s, 4 m/s
(2) 1 m/s, 3 m/s
15mia
3 mie​

Answer 1

Answer:

Av. vel. = 1 m/s

Av. speed = 3 m/s

Explanation:

First of all, we'll find the value of acceleration provided by the force qE on the toy car.

We know that,

    v = u + at

⇒ 6 = 0 + a × 1

⇒ a = 6 m/s²

Now, we'll find the distance traveled by the toy car in first second, through the formula s = ut + $\frac{1}{2}$ at²

It will come out to be 3 meters.

Now, when the direction of electric field is reversed, the toy car will first decelerate with same value of retardation (6m/s²) because the magnitude of electric field is same.

Thence, at t = 2 sec., the car will return to it's initial position (position at t = 0 sec.) and then in third second the car will accelerate towards opposite side, that too with same acceleration (covering same 3 meters in 1 second).

So, total displacement will be 3 meters, and total distance will be 9 meters.

Hence, average velocity will be  $\frac{3}{3}$  m/s = 1 m/s.

And average speed will be  $\frac{9}{3}$  m/s = 3 m/s.