Question

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

Answer 1

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
$\bar v = \dfrac{{at}}{2}$ .