Question

A 1000kg vehicle moving with a speed of 20 m/s is brought to rest in a distance of 50.find the acceleration​

Answer 1

$\textbf{\huge\underline{\underline\red{solution} : - }} \\ \\ \tt \underline\red{given} : - \\ \\ \tt{a \: vehical \: moving \: with \: speed \: of \:} \\ \tt{ 20 \: meter \: per \: second \: and \: stop \:at} \\ \tt{ distanse \: 50 \: meter } \\ \\ \tt \purple{we \: use \: formula \: } \\ \\ \tt \red{ {v}^{2} = {u}^{2} + 2as } \\ \\ \tt \purple{where} \\ \\ \tt \red{v = final \: velocity } \\ \tt \red{u = initial \: velocity } \\ \tt \red{a = acceleration } \\ \tt \red{s = traveling \: distance} \\ \\ \tt \purple{by \:given\: } \\ \tt \red{v = 0} \\ \tt \red{u = 20 \:m/s} \\ \tt \red{s = 50 \: meter} \\ \tt \red{a = \: \: ? } \\ \\ \tt \purple{by \: formula} \\ \\ \tt{ {0}^{2} = {20}^{2} + 2 \times a \times 50 } \\ \\ \tt{0 = 400 + a \times 100} \\ \\ \tt{a \times 100 = - 400} \\ \\ \tt{a = - \frac{400}{100} } \\ \\ \tt{a = - 4 \: m/{s}^{2} } \\ \\ \sf \red{hence \: acceleration \: of \: } \\ \sf\red{vehical \: is \: - 4 \: m/{s}^{2}.} \\ \\ \\ \fcolorbox{red}{aqua}{i \: hope \: it \: helps \: you.}$

Answer 2

$\large{\bf{\red{\underline{\underline{Question:}}}}}$

A 1000kg vehicle moving with a speed of 20 m/s is brought to rest in a distance of 50.find the acceleration?

$\large{\bf{\green{\underline{\underline{Answer:}}}}}$

Given,

• Initial speed = 20m/s
• Distance = 50m

To Find:-

Acceleration.

Formula used:-

${v}^{2} = {u}^{2} + 2as$

V=Final speed

u=initial speed

a=acceleration

s=displacement

Calculations:-

${v}^{2} = {u}^{2} + 2as$

Now substituting the given values:-

${0}^{2} = {20}^{2} + 2 \times a \times 50$

$0 = 400 + a \times 100$

$a \times 100 = - 400$

$a = \frac{ - 400}{100}$

$a = - 4 \frac{m}{ {s}^{2} }$