Question

A body starts from rest with 2/4 uniform acceleration of 2m/s2. Find the distance covered by the body in 2s​

Answer 1

Answer:

Appropriate Question :-

• A body starts from rest with uniform acceleration of 2 m/s². Find the distance covered by the body in 2 seconds.

Given :-

• A body starts from rest with uniform acceleration of 2 m/s².

To Find :-

• What is the distance covered by the body.

Formula Used :-

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}$

where,

• s = Distance Covered
• u = Initial Velocity
• t = Time Taken
• a = Acceleration

Solution :-

Given :

• Initial Velocity (u) = 0 m/s
• Acceleration (a) = 2 m/s²
• Time Taken (t) = 2 seconds

According to the question by using the formula we get,

$\longrightarrow \sf s =\: (0)(2) + \dfrac{1}{2} \times (2)(2)^2$

$\longrightarrow \sf s =\: 0 \times 2 + \dfrac{1}{2} \times 2 \times 2 \times 2$

$\longrightarrow \sf s =\: 0 + \dfrac{1}{2} \times 2 \times 4$

$\longrightarrow \sf s =\: 0 + \dfrac{1}{\cancel{2}} \times {\cancel{8}}$

$\longrightarrow \sf s =\: 0 + 4$

$\longrightarrow \sf\bold{\red{s =\: 4\: m}}$

${\small{\bold{\underline{\therefore\: The\: distance\: covered\: by\: the\: body\: is\: 4\: m\: .}}}}$

Answer 2

$\huge \mathcal {-:Solution:-}$

__________________________________

$\sf \red{\underline{\underline{Given\: data:}}}$

$\dashrightarrow \tt{Acceleration=2m/s^{2}}$

$\dashrightarrow \tt{Time\: taken=2s}$

★As the question says, that a body starts from rest:

$\dashrightarrow \tt{Intial\: Velocity=0}$

__________________________________

$\sf \blue{\underline{\underline{To\: discover:}}}$

★The distance covered by the body in 2s=?

__________________________________

$\sf \pink{\underline{\underline{Kinematic\: equation\: to\: be\: used:}}}$

$\large \implies \tt{S=ut+\frac{1}{2}at^{2}}$

★where,

$\to {S=Distance\: covered\: by\: the\: body}$

$\to {u=Initial\: Velocity}$

$\colon \implies {Zero\: in\: this\: case}$

$\to {a=Acceleration}$

$\to {t=time}$

__________________________________

★Supplanting the values given in the equation,

$\colon \to \tt{S=ut+\frac{1}{2}at^{2}}$

$\colon \to \tt{S=0\times2+\frac{1}{2}\times2\times(2)^{2}}$

$\colon \to \tt{S=0+\frac{1}{2}\times2\times4}$

$\colon \to \tt{S=0+\frac{1}{\cancel{2}}\times{\cancel{2}}\times4}$

$\large \implies \tt \purple {\fbox{S=4m}}$

__________________________________

$\sf \orange{\underline{\underline{Thence,}}}$

★The distance covered by the body in 2s is 4m

__________________________________

$\sf \green{\underline{\underline{ kinematic\: equations:}}}$

• The one which we used in this case to determine distance i.e,

$\dashrightarrow \tt{ S=ut+\frac{1}{2}at^{2}}$

• To determine velocity,

$\dashrightarrow \tt{ v=u+at}$ where, 'v' is the final velociy

• The timeless kinematic equation

$\dashrightarrow \tt{ v^{2}=u^{2}+as}$

• To find the average displacement

$\dashrightarrow \tt{ S=\frac{u+v}{2}\times t}$