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A car travels from rest with 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) $v = \dfrac{{a{t^2}}}{2}$
(B) $v = 2a{t^2}$
(C) $v = \dfrac{{at}}{2}$
(D) None

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Last updated date: 06th Sep 2024
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Answer
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Hint The given problem can be solved by calculating the total distance travelled by car in $t$ seconds. Since the car is moving from rest along a straight road, the initial velocity is taken to be $0$ and the acceleration of the car is uniform at $'a'$. The total distance travelled by car is given by the equation of motion as below.
Formula used: Equation of motion $S = ut + \dfrac{1}{2}a{t^2}$

Complete Step by step solution It is given that the car travels along a straight road. Therefore we will use the equations of motion to find the total distance travelled by car.
Using $S = ut + \dfrac{1}{2}a{t^2}$,
where $S$ is the displacement travelled as the car moves in a straight line,
$u$ is the initial velocity of the car,
$a$ is the acceleration of the car which is given to be $'a'$, and
$t$ is the time taken in seconds for this required calculation.
Since the car travels from rest, the initial velocity of the car is $0$, i.e. $u = 0$.
$ \Rightarrow S = \dfrac{1}{2}a{t^2}$, is the required displacement of the car in $t$ seconds.
Now for finding the average speed of the car in the entire time duration of for $t$ seconds, the required total distance is given by $\dfrac{1}{2}a{t^2}$ as found above. The car travels this distance in $t$ seconds.
We know $Average\,speed = \dfrac{{Total\,displacement}}{{Total\,time\,taken}}$,
Therefore, the average speed
$v = \dfrac{{\dfrac{1}{2}a{t^2}}}{t}$
$ \Rightarrow v = \dfrac{1}{2}at$

Therefore, the correct answer is option (C) $v = \dfrac{{at}}{2}$.

Note In order to find the displacement of the car we have used the equation of motion which can be applied only when the particle (in this case the car) is moving at a constant acceleration. If the acceleration would have been variable, then the differential forms of the equations of motions would have been used.