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 $6m/s$ 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) $1m/s,3m/s$

Step 1,

According to the question

0 < t < 1s : Velocity increases from 0 to 6 m/s

1 < t < 2s : Velocity increases from 6 to 0 m/s but car continues to move forward

2 < t <3s : Since field strength is Same

$\therefore$ Car's velocity increases from 0 to -6 m/s

Step 2,

Distance traveled in 1st second

$S=\left(\frac{u+v}{2}\right)t=\left(\frac{0+6}{2}\right)\left(1\right)=3m$

So according to the moving condition of the car we can say that the total distance traveled is given as

Distance$=9m$

Now,

We know,

Average speed $=\frac{9m}{3s}=3m/s$

And

Displacement $=3m$

So, Average velocity $=\frac{3m}{3s}=1m/s$

Graph related to the above moving condition is given as