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# A car accelerates from $5.0\;m/s$ to $21\;m/s$ at a constant rate of $3.0\;m/s^2$. How far does it travel while accelerating?

Last updated date: 11th Aug 2024
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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:
$v^{2}-u^{2}=2as$

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.

$v=u+at$
$s=ut+\dfrac{1}{2}at^{2}$
$v^{2}-u^{2}=2as$
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 $s$.