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Question Number 191410 by otchereabdullai last updated on 23/Apr/23
 Three villages A, B and C are on a    straight road and B is the mid-way  between A and C. A motor cyclist  moving with a uniform acceleration  passes A, B and C. The speeds with   which the motorcyclist passes A and   C are 20ms^(−1)  and 40ms^−  respectively.  find the speed with which the motor  cyclist passes B.
$$\:{Three}\:{villages}\:{A},\:{B}\:{and}\:{C}\:{are}\:{on}\:{a}\: \\ $$$$\:{straight}\:{road}\:{and}\:{B}\:{is}\:{the}\:{mid}-{way} \\ $$$${between}\:{A}\:{and}\:{C}.\:{A}\:{motor}\:{cyclist} \\ $$$${moving}\:{with}\:{a}\:{uniform}\:{acceleration} \\ $$$${passes}\:{A},\:{B}\:{and}\:{C}.\:{The}\:{speeds}\:{with}\: \\ $$$${which}\:{the}\:{motorcyclist}\:{passes}\:{A}\:{and}\: \\ $$$${C}\:{are}\:\mathrm{20}{ms}^{−\mathrm{1}} \:{and}\:\mathrm{40}{ms}^{−} \:{respectively}. \\ $$$${find}\:{the}\:{speed}\:{with}\:{which}\:{the}\:{motor} \\ $$$${cyclist}\:{passes}\:{B}. \\ $$
Answered by mahdipoor last updated on 23/Apr/23
AB=BC=d  v_C =40        v_A =20      m/s  v_2 ^2 −v_1 ^2 =2ad_(1→2)      (when a=cte)   { ((v_C ^2 −v_B ^2 =2ad)),((v_B ^2 −v_A ^2 =2ad)) :}⇒2v_B ^2 =v_A ^2 +v_C ^2   ⇒v_B =(√((40^2 +20^2 )/2))=20(√(2.5))
$${AB}={BC}={d} \\ $$$${v}_{{C}} =\mathrm{40}\:\:\:\:\:\:\:\:{v}_{{A}} =\mathrm{20}\:\:\:\:\:\:{m}/{s} \\ $$$${v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2}{ad}_{\mathrm{1}\rightarrow\mathrm{2}} \:\:\:\:\:\left({when}\:{a}={cte}\right) \\ $$$$\begin{cases}{{v}_{{C}} ^{\mathrm{2}} −{v}_{{B}} ^{\mathrm{2}} =\mathrm{2}{ad}}\\{{v}_{{B}} ^{\mathrm{2}} −{v}_{{A}} ^{\mathrm{2}} =\mathrm{2}{ad}}\end{cases}\Rightarrow\mathrm{2}{v}_{{B}} ^{\mathrm{2}} ={v}_{{A}} ^{\mathrm{2}} +{v}_{{C}} ^{\mathrm{2}} \\ $$$$\Rightarrow{v}_{{B}} =\sqrt{\frac{\mathrm{40}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{20}\sqrt{\mathrm{2}.\mathrm{5}} \\ $$$$ \\ $$
Commented by otchereabdullai last updated on 23/Apr/23
God bless you sir!
$${God}\:{bless}\:{you}\:{sir}! \\ $$

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