A particle starts from rest and its angular displacement (in radians) is given by θ =(t2/20)+(t/5) . If the angular velocity at the end of t =4 is k, then the value of 5 k is

Question

A particle starts from rest and its angular displacement (in radians) is given by θ =(t2/20)+(t/5) . If the angular velocity at the end of t =4 is k, then the value of 5 k is (A) 0.6 (B) 5 (C) 5 k (D) 3

Tardigrade Admin · Accepted Answer

( d θ/ dt )=(2 t /20)+(1/5) =( t /10)+(1/5) (( d θ/ dt ))ε=4=(4/10)+(1/5) k =(3/5) 5 k =3

Q. A particle starts from rest and its angular displacement (in radians) is given by $θ = \frac{t ^{2}}{20} + \frac{t}{5} .$ If the angular velocity at the end of $t = 4$ is $k$ , then the value of $5 k$ is

$0.6$

5

5 k

3

$\frac{d θ}{d t} = \frac{2 t}{20} + \frac{1}{5}$
$= \frac{t}{10} + \frac{1}{5}$
$(\frac{d θ}{d t})_{ε = 4} = \frac{4}{10} + \frac{1}{5}$
$k = \frac{3}{5}$
$5 k = 3$

Q. A particle starts from rest and its angular displacement (in radians) is given by θ=20t2​+5t​. If the angular velocity at the end of t=4 is k, then the value of 5k is

0.6

5

5 k

3

dtdθ​=202t​+51​ =10t​+51​ (dtdθ​)ε=4​=104​+51​ k=53​ 5k=3

Q. A particle starts from rest and its angular displacement (in radians) is given by $θ = \frac{t ^{2}}{20} + \frac{t}{5} .$ If the angular velocity at the end of $t = 4$ is $k$ , then the value of $5 k$ is

$0.6$

$\frac{d θ}{d t} = \frac{2 t}{20} + \frac{1}{5}$
$= \frac{t}{10} + \frac{1}{5}$
$(\frac{d θ}{d t})_{ε = 4} = \frac{4}{10} + \frac{1}{5}$
$k = \frac{3}{5}$
$5 k = 3$