Dear Student As gievna α ta = ktk is constanta=dvdt= kt∫dv =∫ kt dtv=kt22+CAt t = 0 , speed of particle is = 0, so C = 0So t = nv=kn22Nowv = dxdt =kt22dx = kt22 dt∫dx =∫kt22 dtx =kt33×2+CAt t =0, x = 0 so C= 0x =kt36At t = nx =kn36 Regards

Solve no. 1

Dear Student

A s g i e v n a α t a = k t k i s c o n s \tan t a = \frac{d v}{d t} = k t \int d v = \int k t d t v = [\frac{k t^{2}}{2} + C] A t t = 0, s p e e d o f p a r t i c l e i s = 0, s o C = 0 S o t = n v = \frac{k n^{2}}{2} N o w v = \frac{d x}{d t} = \frac{k t^{2}}{2} d x = \frac{k t^{2}}{2} d t \int d x = \int \frac{k t^{2}}{2} d t x = k \frac{t^{3}}{3 \times 2} + C A t t = 0, x = 0 s o C = 0 x = k \frac{t^{3}}{6} A t t = n x = k \frac{n^{3}}{6}

Regards

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