Given-that-the-velocity-v-of-a-body-t-seconds-after-passing-a-point-O-is-found-by-v-2-1-k-P-P-kv-o-2-e-2kt-m-determine-the-distance-covered-by-the-body-one-hour-after-passing-

Question Number 896 by 112358 last updated on 15/Apr/15

G i v e n t h a t t h e v e l o c i t y v o f a b o d y

t s e c o n d s a f t e r p a s s i n g a p o i n t O

i s f o u n d b y

v^{2} = \frac{1}{k} [P - (P - {k v}_{o}^{2}) e^{- \frac{2 k t}{m}}]

d e t e r m i n e t h e d i s t a n c e c o v e r e d

b y t h e b o d y o n e h o u r a f t e r

p a s s i n g O . T h e b o d y m o v e s a l o n g

a s t r a i g h t l i n e . k, P, v_{o} a n d m

a r e c o n s t a n t s a n d t i s m e a s u r e d i n

s e c o n d s .

Commented by 123456 last updated on 15/Apr/15

\underset{0}{\int^{3600}} \sqrt{\frac{1}{k} [P - (P - {k v}_{0}^{2}) e^{- \frac{2 k t}{m}}]} d t

\underset{0}{\int^{3600}} \sqrt{\frac{P}{k} - \frac{P - {k v}_{0}^{2}}{k} e^{- \frac{2 k t}{m}}} d t

\underset{0}{\int^{3600}} \sqrt{α + β e^{γ t}} d t

α = \frac{P}{k}

β = \frac{{k v}_{0}^{2} - P}{k}

γ = - \frac{2 k}{m}

Commented by prakash jain last updated on 16/Apr/15

\int \sqrt{α + β e^{γ x}} d x = \frac{2 (\sqrt{α + β e^{γ x}} - \sqrt{α} \tanh^{- 1} (\frac{\sqrt{α + β e^{γ x}}}{\sqrt{α}}))}{γ}