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Direction: A particle initially (i.e., at time t=0) moving with a velocity u subjected to a retarding force, as a result of which it decelerates at a rate a=−kv√ where v is the instantaneous velocity and k is a positive constant.\displaystyle{T}{h}{e}{d}{i}{s}{\tan{{c}}}{e}{c}{o}{v}{e}{r}{e}{d}{b}{y}{t}{h}{e}{p}{a}{r}{t}{i}{c}\le{b}{e}{f}{\quad\text{or}\quad}{e}{c}{o}\min{g}\to{r}{e}{s}{t}{i}{s}
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[d] To find the distance s covered in this time, we usev1/2= u1/2kt2v=u−ktu1/2+k2t24Therefore, dsdt=u−ktu1/2+k2t24t=0 to t=τ we haveor s−uτ−12ku1/2τ2+112k2τ3 ... (iv)s=2u3/2k−4u3/22k+8u3/212k\displaystyle{\quad\text{or}\quad}{s}={2}{u}\frac{{3}}{{23}}{k}
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Question Text | Direction: A particle initially (i.e., at time t=0) moving with a velocity u subjected to a retarding force, as a result of which it decelerates at a rate a=−kv√ where v is the instantaneous velocity and k is a positive constant.\displaystyle{T}{h}{e}{d}{i}{s}{\tan{{c}}}{e}{c}{o}{v}{e}{r}{e}{d}{b}{y}{t}{h}{e}{p}{a}{r}{t}{i}{c}\le{b}{e}{f}{\quad\text{or}\quad}{e}{c}{o}\min{g}\to{r}{e}{s}{t}{i}{s} |
Answer Type | Text solution:1 |
Upvotes | 150 |