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 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

Hint: Divide the time taken in different phases and then calculate the distance using it and thus the speed and velocity. The area under the velocity-time curve gives the displacement of the object in the given time interval.

Complete answer:

According to the question we can see that from,

\[0\text{ }<\text{ }t\text{ }<\text{ }1s\]:  Velocity increases from 0 to \[6\text{ }m/s\]. 

\[1\text{ }<\text{ }t\text{ }<\text{ }2s\]: Velocity decreases from \[6\text{ }to\text{ }0\text{ }m/s\] but car continues to move forward 

\[2\text{ }<\text{ }t\text{ }<3s\]: Since, field strength is same 

Therefore, we see that, Car's velocity decreases from \[0\text{ to }-\text{ }6\text{ }m/s\] 

 The distance traveled in 1st second: 

Is the area that has been covered in the time interval \[0<t<1s\], which is:

$ \text{Area of triangle}=\dfrac{1}{2}\times \text{Base}\times \text{Height}$

$ {{A}_{1}}=\dfrac{1}{2}\times 1\times 6=3m $

The distance traveled in 2nd second is the area that has been covered in the time interval \[\text{1}<t<\text{2}s\], which is:

\[{{A}_{2}}=\dfrac{1}{2}\times 1\times 6=3m\]

The distance traveled in 3rd second is the area that has been covered in the time interval \[2<t<3s\], which is:

\[{{A}_{3}}=\dfrac{1}{2}\times 1\times (-6)=-3m\]

The negative sign suggests that the area formed is under the x axis.

Now we get that, Distance is the length that has been totally travelled. Therefore:

$\text{Distance in 3 seconds }=\left| {{A}_{1}} \right|+\left| {{A}_{2}} \right|+\left| {{A}_{3}} \right|=3+3+3=9m$ 

So, the average speed becomes:

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

Similarly, Displacement is the change in position from the initial and the final point. 

$\therefore$ $\text{Displacement in 3 seconds }={{A}_{1}}+{{A}_{2}}+{{A}_{3}}=3+3-3=3m$ 

So, the average velocity becomes \[\text{Average velocity }=\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{3m}{3s}=1m/s\]

Therefore, the correct answer is option C.

Note:
The frictionless surface and charge q given in the question are not of much use. Such things are used to confuse the students which sometimes lead them to go through the wrong approach. So, don’t be confused with such things since it’s not important that all the details given in question are to be used for evaluation of answers. Similarly, the charge q and the field E do not provide any help in solving the question.