Question

The velocity of a body changes from 5
m/s to 20 m/s in 5 s. calculate the
acceleration of the body. Also find the
distance travelled by the body in 5 s.​

whoever gives the correct answer I will mark them as the brainliest

Answer 1

$\small \underline \bold{For \: a \: body -}$

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

$\: \: \: \:$$\sf{Initial \: velocity \: (u) = 5 \: m/s}$

$\: \: \: \:$$\sf{Final \: Velocity \: (v) = 20 \: m/s}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$$\sf{Time \: (t) = 5 \: Second}$

$\large \underline \bold{To \: Find :-}$

$\sf{Calculate \: the \: acceleration \: of \: the \: body \: and \: the \: distance}$

$\: \: \:$$\sf{travelled \: by \: the \: body \: in \: 5 \: second \: ?}$

$\large \underline \bold{Using \: Formulas -}$

$\: \: \: \: \: \:$$\sf{a = \dfrac{(v - u)}{t}}$

$\: \: \: \: \: \:$$\sf{S = ut + \dfrac{1}{2}at^{2}}$

$\sf{Here \: ,}$

$\: \: \: \: \: \:$$\sf{a = acceleration \: (in \: m/s^{2})}$

$\: \: \: \: \: \:$$\sf{u = initial \: velocity \: (in \: m/s)}$

$\: \: \: \: \: \:$$\sf{v = final \: velocity \: (in \: m/s)}$

$\: \: \: \: \: \:$$\sf{t = time \: (in \: second)}$

$\: \: \: \: \: \:$$\sf{S = travelled \: distance \: (in \: metre)}$

$\large \underline \bold{Solution :-}$

$\sf{Firstly \: -}$

$\: \: \: \: \: \:$$\sf{a =\dfrac{(20 - 5)}{5}}$

$\: \: \: \: \: \:$$\sf{a =\dfrac{15}{5}}$

$\: \: \: \: \: \:$$\small \bold{a = 3 \: m/s^{2}}$

$\sf{Now \: ,}$

$\small \underline \bold{Distance \: travelled \: by \: the \: body -}$

$\: \: \:$$\sf{S = 5(5) + \dfrac{1}{2}(3)(5)^{2}}$

$\: \: \:$$\sf{S = 25 + \dfrac{1}{2}(3)(25)}$

$\: \: \:$$\sf{S = 25 + \dfrac{75}{2}}$

$\: \: \:$$\sf{S = 25 + 37.5}$

$\: \: \:$$\small \bold{S = 62.5 \: metre}$