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Question Number 181052 by Acem last updated on 20/Nov/22

Commented by Acem last updated on 20/Nov/22

∗ When the car is 6 Km from A

$$\ast\:{When}\:{the}\:{car}\:{is}\:\mathrm{6}\:{Km}\:{from}\:{A} \\ $$

Commented by mr W last updated on 22/Nov/22

((220)/( (√7))) km/h  what′s your solution?

$$\frac{\mathrm{220}}{\:\sqrt{\mathrm{7}}}\:{km}/{h} \\ $$$${what}'{s}\:{your}\:{solution}? \\ $$

Commented by Acem last updated on 22/Nov/22

The same result (:  let′s write our methods

$${The}\:{same}\:{result}\:\left(:\right. \\ $$$${let}'{s}\:{write}\:{our}\:{methods} \\ $$

Answered by Acem last updated on 22/Nov/22

Answered by mr W last updated on 23/Nov/22

Commented by mr W last updated on 23/Nov/22

L=(√(9^2 +18^2 +9×18))=9(√7)  cos θ=((9^2 +(9(√7))^2 −18^2 )/(2×9×9(√7)))=(2/( (√7)))  since “I” don′t move, how fast the  distance from the car to me changes  means how fast the car moves   towards me at that moment,  that′s   the component of its velocity in my   direction, i.e.  v cos θ=((2v)/( (√7)))=((220)/( (√7))) km/h.

$${L}=\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{18}^{\mathrm{2}} +\mathrm{9}×\mathrm{18}}=\mathrm{9}\sqrt{\mathrm{7}} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{9}^{\mathrm{2}} +\left(\mathrm{9}\sqrt{\mathrm{7}}\right)^{\mathrm{2}} −\mathrm{18}^{\mathrm{2}} }{\mathrm{2}×\mathrm{9}×\mathrm{9}\sqrt{\mathrm{7}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{7}}} \\ $$$${since}\:``{I}''\:{don}'{t}\:{move},\:{how}\:{fast}\:{the} \\ $$$${distance}\:{from}\:{the}\:{car}\:{to}\:{me}\:{changes} \\ $$$${means}\:{how}\:{fast}\:{the}\:{car}\:{moves}\: \\ $$$${towards}\:{me}\:{at}\:{that}\:{moment},\:\:{that}'{s}\: \\ $$$${the}\:{component}\:{of}\:{its}\:{velocity}\:{in}\:{my}\: \\ $$$${direction},\:{i}.{e}. \\ $$$${v}\:\mathrm{cos}\:\theta=\frac{\mathrm{2}{v}}{\:\sqrt{\mathrm{7}}}=\frac{\mathrm{220}}{\:\sqrt{\mathrm{7}}}\:{km}/{h}. \\ $$

Commented by mr W last updated on 23/Nov/22

if “I” also move with speed v_1 , then  it is v cos θ+v_1  cos θ_1 .

$${if}\:``{I}''\:{also}\:{move}\:{with}\:{speed}\:{v}_{\mathrm{1}} ,\:{then} \\ $$$${it}\:{is}\:{v}\:\mathrm{cos}\:\theta+{v}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} . \\ $$

Commented by mr W last updated on 23/Nov/22

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