Solve this-Q- nbsp- nbsp-A particle starts its motion from rest and moves with constant acceleration a1 nbsp-for time tl and then nbsp-it retards with constant rate a2 for time t2 until it nbsp-comes rest- Then the ratio of average speed nbsp-and maximum Speed during the complete motion nbsp-will be nbsp-a- a1t2-a2t1-b- a1t1-a2t2-c- 1-2-d- 1-1

Dear Student, maximum #160;speed #160;v=a1t1 #160; #160; #160;maximum #160;speed.S1=a1t122S2=a1t1t2 a2t222average #160;speed #160;=a1t122+a1t1t ...

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Byju's Answer

Standard XII

Physics

Relating Circular Motion to SHM

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Question

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Q. A particle starts its motion from rest and moves with constant acceleration a₁ for time t_l and then it retards with constant rate a₂ for time t₂ until it comes rest, Then the ratio of average speed and maximum Speed during the complete motion will be
(a) a₁t₂:a₂t₁
(b) a₁t₁:a₂t₂
(c) 1:2
(d) 1:1

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Solution

Dear Student,

$m a x i m u m s p e e d v = a_{1} t_{1} m a x i m u m s p e e d . S_{1} = \frac{a_{1} {t_{1}}^{2}}{2} S_{2} = a_{1} t_{1} t_{2} - \frac{a_{2} {t_{2}}^{2}}{2} a v e r a g e s p e e d = \frac{\frac{a_{1} {t_{1}}^{2}}{2} + a_{1} t_{1} t_{2} - \frac{a_{2} {t_{2}}^{2}}{2}}{t_{1} + t_{2}} r a t i o = \frac{\frac{\frac{a_{1} {t_{1}}^{2}}{2} + a_{1} t_{1} t_{2} - \frac{a_{2} {t_{2}}^{2}}{2}}{t_{1} + t_{2}}}{a_{1} t_{1}} = \frac{\frac{a_{1} {t_{1}}^{2}}{2} + a_{1} t_{1} t_{2} - \frac{a_{2} {t_{2}}^{2}}{2}}{(a_{1} t_{1}) (t_{1} + t_{2})} = 1 : 2 R e g a r d s$

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Dear Student, maximum speed v=a1t1 maximum speed.S1=a1t122S2=a1t1t2-a2t222average speed =a1t122+a1t1t2-a2t222t1+t2ratio =a1t122+a1t1t2-a2t222t1+t2a1t1=a1t122+a1t1t2-a2t222a1t1t1+t2=1:2Regards