A-particle-starts-from-rest-with-acceleration-30-6t-ms-2-at-time-t-Where-will-the-particle-come-to-rest-again-

Question Number 43354 by pieroo last updated on 10/Sep/18

A particle starts from rest with acceleration (30 + 6 t)

{ms}^{- 2} at time t . Where will the particle come to rest

again ?

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

\frac{d v}{d t} = 30 + 6 t

h e r e a c c e l a r a t i o n i s + v e s o n o q u e s t i o n

a r i s e t h a t p a r t c l e w i l l c o m e t o r e s t a g a i n

d v = (30 + 6 t) d t

v = 30 t + 3 t^{2} + c

t = 0 v = 0 s o c = 0

v = 30 t + 3 t^{2} i f p a r t i c l e a t r e s t v = 0

30 t + 3 t^{2} = 0

3 t (10 + t) = 0

t = 0 b u t t c a n n o t b e n e g e t i v e (t = - 10)

s o p a r t i c l e c a n n e v e r c o m e t o r e s t a g a i n \dots