A car travels from rest with a constant acceleration $ 'a' $ for $ t $ seconds. What is the average speed of the car for its journey, if the car moves along a straight road?
(A) $ \dfrac{{a{t^2}}}{2} $
(B) $ 2a{t^2} $
(C) $ \dfrac{{at}}{2} $
(D) None

Hint : Since the car moves on a straight road, then the equation of motion on a straight line can be used. The average velocity is the total displacement covered within a range of time divided by that elapsed time. Since it was at rest, the initial velocity should be taken as zero.

Formula used: In this solution we will be using the following formula;
 $ \bar v = \dfrac{s}{t} $ where $ v $ is the average velocity, $ s $ is the displacement, and $ t $ is the time taken to cover the displacement.
 $ s = ut + \dfrac{1}{2}a{t^2} $ where $ u $ is the initial velocity, and $ a $ is the acceleration. $ v = u + at $ , where $ v $ is the final velocity.
 $ \bar v = \dfrac{{v + u}}{2} $ , where $ v $ is the final velocity, and $ u $ is the initial velocity of a body.

Complete step by step answer
The equation of motion of a body moving in constant acceleration is given as
 $ s = ut + \dfrac{1}{2}a{t^2} $ where $ s $ is the displacement $ u $ is the initial velocity, $ t $ is time, and $ a $ is the acceleration.
To calculate the average velocity, we recall that it is given by
 $ \bar v = \dfrac{s}{t} $
Hence, on substitution of the displacement from $ s = ut + \dfrac{1}{2}a{t^2} $ we have
 $ \bar v = \dfrac{{ut + \dfrac{1}{2}a{t^2}}}{t} $ . Hence, by dividing both numerator and denominator by $ t $ , we have
 $ \bar v = u + \dfrac{1}{2}at $ ,
The car in the question is said to start from rest, hence, $ u = 0 $ , then
 $ \bar v = \dfrac{1}{2}at = \dfrac{{at}}{2} $
Hence, the correct answer is C.

Note
Alternatively, for a constantly accelerated motion, the average velocity can be given as
 $ \bar v = \dfrac{{v + u}}{2} $ where $ v $ is the final velocity, and $ u $ is the initial velocity of a body.
Now, the velocity after any time $ t $ (final velocity) of a motion on a straight line can be given as $ v = u + at $ . Hence, on inserting this into the above equation, we have that
 $ \bar v = \dfrac{{u + at + u}}{2} = \dfrac{{2u + at}}{2} $
since, $ u = 0 $ , then
 $ \bar v = \dfrac{{at}}{2} $ .