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Question-14246




Question Number 14246 by tawa tawa last updated on 30/May/17
Commented by ajfour last updated on 30/May/17
Commented by ajfour last updated on 30/May/17
a=((v−u)/(Δt)) =((25−30)/3)=−(5/3)m/s^2   let total time to stop be T, then  from v=u+at  0=30−(5/3)T    ⇒ T=18s  and from:  s=ut+(1/2)at^2    distance=30×18−((1/2))((5/3))(18)^2                      =540−270                      =270m .
$${a}=\frac{{v}−{u}}{\Delta{t}}\:=\frac{\mathrm{25}−\mathrm{30}}{\mathrm{3}}=−\frac{\mathrm{5}}{\mathrm{3}}{m}/{s}^{\mathrm{2}} \\ $$$${let}\:{total}\:{time}\:{to}\:{stop}\:{be}\:{T},\:{then} \\ $$$${from}\:\boldsymbol{{v}}=\boldsymbol{{u}}+\boldsymbol{{at}} \\ $$$$\mathrm{0}=\mathrm{30}−\frac{\mathrm{5}}{\mathrm{3}}{T}\:\:\:\:\Rightarrow\:{T}=\mathrm{18}{s} \\ $$$${and}\:{from}:\:\:\boldsymbol{{s}}=\boldsymbol{{ut}}+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{at}}^{\mathrm{2}} \\ $$$$\:{distance}=\mathrm{30}×\mathrm{18}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{5}}{\mathrm{3}}\right)\left(\mathrm{18}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{540}−\mathrm{270}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{270}{m}\:. \\ $$
Commented by tawa tawa last updated on 30/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by ajfour last updated on 30/May/17
Commented by ajfour last updated on 30/May/17
(2A)^2 +A^2 =64  5A^2 =64  A=(8/( (√5)))  .
$$\left(\mathrm{2}{A}\right)^{\mathrm{2}} +{A}^{\mathrm{2}} =\mathrm{64} \\ $$$$\mathrm{5}{A}^{\mathrm{2}} =\mathrm{64} \\ $$$${A}=\frac{\mathrm{8}}{\:\sqrt{\mathrm{5}}}\:\:. \\ $$
Commented by tawa tawa last updated on 30/May/17
Thanks for everytime. God bless you sir.
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{everytime}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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