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**Hint:**We know that acceleration is the rate of change of velocity with respect to time. Here the acceleration, initial and final speed of the car is given, using the equation of motion for a straight line, we can find the distance covered by the car.

**Formula used:**

$v^{2}-u^{2}=2as$

**Complete answer:**

We know that the acceleration is the rate of change of velocity with respect to time, it is mathematically given as, $a=\dfrac{dv}{dt}$, where $a$ is the acceleration due to the change in velocity $d\;v$ during time $d\;t$ . Also, acceleration is a vector quantity which has both direction and magnitude. From Newton’s laws of motion we know that a body undergoes acceleration when an external force is applied on the body and given as $F=ma$ where, $F$ is the force, $m$ is mass and $a$ is acceleration.

From the equation of motion we know that $v^{2}-u^{2}=2as$

On rearranging we get,

$\implies 2as=v^{2}-u^{2}$

$\implies s=\dfrac{v^{2}-u^{2}}{2a}$

Given that $v=21m/s$ , $u=5m/s$ and $a=3m/s^{2}$

Substituting the values, we get

$\implies s= \dfrac{(21)^{2}-(5)^{2}}{2\times 3}$

$\implies s=\dfrac{441-25}{6}$

$\implies s=\dfrac{416}{6}$

$\therefore s=69.34m$

Thus the car covers a distance of $69.34\;m$ during its acceleration.

**Additional information:**

we know that when an object travels in a straight line, then the following are its equation of motions:

$v=u+at$

$s=ut+\dfrac{1}{2}at^{2}$

$v^{2}-u^{2}=2as$

Where $v$ is the final velocity, $u$ is the initial velocity ,$a$ is the acceleration during the time $t$ and the object covers a distance $s$.

**Note:**

The magnitude of acceleration can be both positive and negative. Negative acceleration is also known as deceleration. Since acceleration is dependent on the direction of motion, it can further be classified into linear, which is due to motion in a straight line and radial speed, which is due to circular motion. But speed is only positive, as it’s a scalar quantity which has magnitude and no direction.

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