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# The magnitude of its acceleration (in $m/{s^2}$).

Last updated date: 22nd Jul 2024
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Hint: We must know first about the basics. Acceleration is defined as the time of rate of change of velocity. Since velocity has both magnitude and direction, so does acceleration. The length of the vector is magnitude. Its direction is the direction of the vector. So, the magnitude of the acceleration is the magnitude of the acceleration vector while the direction of the acceleration is the direction of the acceleration vector.

According to Newton’s second law, the acceleration (a) is proportional to the net force (F), and inversely proportional to the object’s mass (m). By analogy: the magnitude of the acceleration is proportional to the magnitude of this force. Therefore, the formula regarding this is: $\left| a \right| = \left| F \right|/m$.
${v^2} = {v^2} + 2as$
U= initial speed
V=final speed
S= distance covered
a= acceleration
It is given that,
$u = 20 \\ v = 30 \\ s = 200 \\$
So,
${\left( {30} \right)^2} = {\left( {20} \right)^2} + 2a \times 200 \\ \Rightarrow 900 = 400 + 400a \\ \Rightarrow a = \dfrac{{500}}{{400}} \\ \Rightarrow a = 1.25m/{s^2} \\$
Therefore, the correct answer is : $a = 1.25m/{s^2}$

Note: Acceleration, is a vector, that means it has both magnitude and direction. Therefore, the magnitude only describes part of any accelerated motion. The SI unit for the acceleration is : meter per second squared. In mechanics, acceleration is the rate of change of the velocity of an object with respect to time.