A particle moves so that its acceleration is always twice its velocity. If its initial velocity is 0.1 ms -1, its velocity after it has gone 0.1 m is

Question

A particle moves so that its acceleration is always twice its velocity. If its initial velocity is 0.1 ms -1, its velocity after it has gone 0.1 m is (A) 0.3 ms -1 (B) 0.7 ms -1 (C) 1.2 ms -1 (D) 3.6 ms -1

Tardigrade Admin · Accepted Answer

(d v/d t)=2 v ⇒ (d v/d t)=(2 d x/d t) d v=2 d x ∫0.1v d v=2 ∫00.1 d x (y-0.1)=2(0.1) ⇒ v=0.3 ms -1

Q. A particle moves so that its acceleration is always twice its velocity. If its initial velocity is $0.1 m s^{- 1},$ its velocity after it has gone $0.1 m$ is

$0.3 m s^{- 1}$

$0.7 m s^{- 1}$

$1.2 m s^{- 1}$

$3.6 m s^{- 1}$

$\frac{d v}{d t} = 2 v \Rightarrow \frac{d v}{d t} = \frac{2 d x}{d t}$
$d v = 2 d x$
$\int_{0.1}^{v} d v = 2 \int_{0}^{0.1} d x$
$(y - 0.1) = 2 (0.1) \Rightarrow v = 0.3 m s^{- 1}$

Q. A particle moves so that its acceleration is always twice its velocity. If its initial velocity is 0.1ms−1, its velocity after it has gone 0.1m is

0.3ms−1

0.7ms−1

1.2ms−1

3.6ms−1