Question

if a constant force acts on a body initially kept at rest,the distance moved by body in time to is proportional to​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

$\huge{\underline{\underline{\sf{Given : -}}}}$

The body is at rest

• Initial Velocity of the body is zero » u = 0 m/s

• A constant force acts on the body

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We Know that,

$\huge{\boxed{\boxed{\tt{F = ma}}}}$

$\large{\leadsto \red{\sf{a = \frac{F}{m}}}...................(1)}$

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Using the Kinetic Equation,

$\huge{\boxed{\boxed{\mathtt{s = ut + \frac{1}{2}a{t}^{2}}}}}$

$\large{\longrightarrow \ \sf{s = \frac{1}{2}a{t}^{2}}}$

From (1),we obtain :

$\\ \\ \longrightarrow \ \sf{s = \frac{F}{2m}{t}^{2}}$

$\large{\longrightarrow \ \sf{{t}^{2} = \frac{2ms}{F}}}$

Since,

Mass of the object is constant » It's magnitude will be considered as one.

$\large{\longrightarrow \ \sf{{t}^{2} \propto \frac{s}{F}}}$

$\huge{\longrightarrow \ \underline{\boxed{\green{\mathtt{s \propto F{t}^{2}}}}}}$

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Here,

Distance covered is

• Directly proportional to the product of force applied and square of time taken

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Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Initial velocity (u) = 0 m/s.
• Let the constant Force be "F".
• And Time be "t".

$\huge{\bold{\underline{Explanation:-}}}$

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From Newton's second law of motion,

$\large{\boxed{\tt F = ma}}$

Now, Rearranging,

$\large{\tt \leadsto F = m \times a}$

$\large{\tt \leadsto a = \dfrac{F}{m}}$

$\large{\tt \leadsto a = \dfrac{F}{m} \: ------(1)}$

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Now, From second kinematic equation,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

As initial velocity is zero.

$\large{\tt \leadsto S = 0 \times t + \dfrac{1}{2}at^2}$

$\large{\tt \leadsto S = 0 + \dfrac{1}{2}at^2}$

$\large{\tt \leadsto S = \dfrac{1}{2}at^2}$

Substituting the value of acceleration From equation (1).

$\large{\tt \leadsto S = \dfrac{1}{2}t^2 \times a}$

$\large{\tt \leadsto S = \dfrac{1}{2}t^2 \times \dfrac{F}{m}}$

$\large{\tt \leadsto S = \dfrac{F \times t^2}{2m}}$

$\large{\tt \leadsto S = \dfrac{F t^2}{2m}}$

since, Here mass and the numeral number 2 is constant

Therefore , it becomes,

$\huge{\boxed{\boxed{\tt S \propto Ft^2}}}$

Therefore, The distance moved by the body is proportional to Force applied and Square of the time taken.

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