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Question Number 15856 by Tinkutara last updated on 14/Jun/17
A particle is moving along a straight  line with uniform acceleration has  velocities 7 m/s at P and 17 m/s at Q.  If R is the midpoint of PQ, then the  average velocity between P and R is?
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{line}\:\mathrm{with}\:\mathrm{uniform}\:\mathrm{acceleration}\:\mathrm{has} \\ $$$$\mathrm{velocities}\:\mathrm{7}\:\mathrm{m}/\mathrm{s}\:\mathrm{at}\:{P}\:\mathrm{and}\:\mathrm{17}\:\mathrm{m}/\mathrm{s}\:\mathrm{at}\:{Q}. \\ $$$$\mathrm{If}\:{R}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{PQ},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{average}\:\mathrm{velocity}\:\mathrm{between}\:{P}\:\mathrm{and}\:{R}\:\mathrm{is}? \\ $$
Answered by ajfour last updated on 14/Jun/17
(v_(PR) )_(avg) =(v_P /2)(1+(1/( (√2)))(√(1+(v_Q ^2 /v_P ^2 ))) ) .
$$\left({v}_{{PR}} \right)_{{avg}} =\frac{{v}_{{P}} }{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{\mathrm{1}+\frac{{v}_{{Q}} ^{\mathrm{2}} }{{v}_{{P}} ^{\mathrm{2}} }}\:\right)\:. \\ $$
Commented by Tinkutara last updated on 14/Jun/17
Can you tell the final answer please?
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{tell}\:\mathrm{the}\:\mathrm{final}\:\mathrm{answer}\:\mathrm{please}? \\ $$
Commented by ajfour last updated on 14/Jun/17
   10m/s .
$$\:\:\:\mathrm{10}{m}/{s}\:. \\ $$
Commented by Tinkutara last updated on 14/Jun/17
Do you mean that for calculating this,  the formula is v_(av)  between P and R =  ((v_P  + v_R )/2) ? I have calculated v_R .
$$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{that}\:\mathrm{for}\:\mathrm{calculating}\:\mathrm{this}, \\ $$$$\mathrm{the}\:\mathrm{formula}\:\mathrm{is}\:{v}_{\mathrm{av}} \:\mathrm{between}\:{P}\:\mathrm{and}\:{R}\:= \\ $$$$\frac{{v}_{{P}} \:+\:{v}_{{R}} }{\mathrm{2}}\:?\:\mathrm{I}\:\mathrm{have}\:\mathrm{calculated}\:{v}_{{R}} . \\ $$
Commented by ajfour last updated on 14/Jun/17
v_R =(√((v_P ^2 +v_Q ^2 )/2))   =13m/s .
$${v}_{{R}} =\sqrt{\frac{{v}_{{P}} ^{\mathrm{2}} +{v}_{{Q}} ^{\mathrm{2}} }{\mathrm{2}}}\:\:\:=\mathrm{13}{m}/{s}\:. \\ $$

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