A-particle-starts-from-rest-and-moves-in-a-straight-line-on-a-smooth-horizontal-surface-Its-acceleration-at-time-t-seconds-is-given-by-k-4v-1-ms-2-

Question Number 91149 by Rio Michael last updated on 28/Apr/20

A particle starts from rest and moves in a straight line on a smooth

horizontal surface . Its acceleration at time t seconds is given by

k (4 v + 1) {ms}^{- 2}

where k is a positve constant and v {ms}^{- 1} is the speed of the particle .

Given that v = \frac{e^{2} - 1}{4} when t = 1 . show that

v = \frac{1}{4} (e^{2 t} - 1)

Commented by mr W last updated on 28/Apr/20

a n s w e r g i v e n i s w r o n g

Answered by mr W last updated on 28/Apr/20

a = \frac{d v}{d t} = k (4 v + 1)

\frac{d v}{4 v + 1} = k d t

\int \frac{d v}{4 v + 1} = k \int d t

\frac{1}{4} \ln (4 v + 1) = k t + C

\frac{1}{4} \ln (4 \times \frac{e^{2} - 1}{4} + 1) = k + C

\frac{1}{4} \ln (e^{2}) = k + C

\frac{1}{4} [\ln (4 v + 1) - \ln (e^{2})] = k (t - 1)

(4 v + 1) e^{- 2} = e^{4 k (t - 1)}

\Rightarrow v = \frac{1}{4} [e^{4 k (t - 1) + 2} - 1]

Commented by Rio Michael last updated on 28/Apr/20

thank you sir, but i have some doubts,

first sir : at t = 1, \frac{1}{4} \ln (4 v + 1) = k t + C is suppose to equal

\frac{1}{4} \ln (4 v + 1) = k + C {don}^{'} t understand why it equals C .

also, \frac{1}{4} \ln (e^{2}) = k + C \Rightarrow \frac{1}{4} \ln (4 v + 1) = k + C

{how}^{'} d you get \frac{1}{4} [\ln (4 v + 1) - \ln (e^{2})] = k t ? ? ?

Commented by Rio Michael last updated on 28/Apr/20

this is my approach sir, ofcourse from your method .

at rest v = 0

a = k (4 v + 1) {ms}^{- 2}, k > 0

a = \frac{d v}{d t}

\Rightarrow \frac{d v}{d t} = k (4 v + 1)

\frac{d v}{4 v + 1} = k d t \Rightarrow \int \frac{d v}{4 v + 1} = k \int d t

\frac{1}{4} \ln (4 v + 1) = k t + C

at v = 0, t = 0

\Rightarrow \frac{1}{4} \ln (4 (0) + 1) = C

\Rightarrow C = 0

now t = 1 \Rightarrow v = \frac{e^{2} - 1}{4}

so \frac{1}{4} \ln [4 (\frac{e^{2} - 1}{4}) + 1] = k \Rightarrow k = 2

\frac{1}{4} \ln (4 v + 1) = 2 k \Rightarrow v = \frac{1}{4} (e^{8 t} - 1)