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Average velocity of particle moving in a straight line with constant acceleration and initial velocity u in first t seconds is?
A. $u + \dfrac{1}{2}at$
B. $u + at$
C. $\dfrac{{u + at}}{2}$
D. $\dfrac{u}{2}$

Last updated date: 13th Jun 2024
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Hint:The average velocity of an object is its total displacement divided by the total time taken. In other words, it is the rate at which an object changes its position from one place to another. Average velocity is a vector quantity.So, we can find the displacement in time t using the second equation of motion for time t and calculate the average velocity.

Complete step by step answer:
According to the question, we are given a particle with
Initial velocity= u
Acceleration= a
Time taken= t
Let the displacement of the particle in time t seconds be s. Then, according to the second equation of motion, we can write
\[s = ut + \dfrac{1}{2}a{t^2}\]
Now, we know that average velocity is given by the total displacement divided by the time taken to cover it. Let the average velocity be v. Then,
v = \dfrac{s}{t} \\
\Rightarrow v = \dfrac{{ut + \dfrac{1}{2}a{t^2}}}{t} \\
\therefore v = u + \dfrac{1}{2}at \\
Therefore, the correct answer is option A.

Note:We are able to use the equations of motion because the equations of motion are derived considering the acceleration to be constant. We could also have used the third equation of motion but it consists of a velocity term so it is of no use here. Also, we shouldn’t confuse average speed with average velocity. Average speed doesn't equal the magnitude of the common velocity. People might imagine that average speed and average velocity are just different names for the identical quantity, but average speed depends on distance and average velocity depends on displacement. If an object changes direction in its journey, then the typical speed is greater than the magnitude of the common velocity.