Velocity v of a particle moving along x axis as a function of time is given by v = 2t m/s. Initially the particle is to the right of the origin and 2 m away from it. Find the position (distance from origin) of the particle after first 3 s

Question

Velocity v of a particle moving along x axis as a function of time is given by v = 2t m/s. Initially the particle is to the right of the origin and 2 m away from it. Find the position (distance from origin) of the particle after first 3 s (A) 5 m (B) 7 m (C) 11 m (D) 9 m

Tardigrade Admin · Accepted Answer

v = 2t ∫ limitszx dx = ∫ limits03 v dt ⇒ x - 2 = ∫ limits03 2t dt ⇒ x - 2 = [(2t2/2)]03 ⇒ x = 11 m

Q. Velocity $v$ of a particle moving along x axis as a function of time is given by $v = 2 t m / s$ . Initially the particle is to the right of the origin and $2 m$ away from it. Find the position (distance from origin) of the particle after first $3 s$

5 m

7 m

11 m

9 m

$v = 2 t$

$z \int x d x = 0 \int 3 v d t$

$\Rightarrow x - 2 = 0 \int 3 2 t d t$

$\Rightarrow x - 2 = [\frac{2 t ^{2}}{2}]_{0}^{3}$

$\Rightarrow x = 11 m$

Q. Velocity v of a particle moving along x axis as a function of time is given by v=2tm/s. Initially the particle is to the right of the origin and 2m away from it. Find the position (distance from origin) of the particle after first 3s

5 m

7 m

11 m

9 m

v=2t z∫x​dx=0∫3​vdt ⇒x−2=0∫3​2tdt ⇒x−2=[22t2​]03​ ⇒x=11m

Q. Velocity $v$ of a particle moving along x axis as a function of time is given by $v = 2 t m / s$ . Initially the particle is to the right of the origin and $2 m$ away from it. Find the position (distance from origin) of the particle after first $3 s$

$v = 2 t$

$z \int x d x = 0 \int 3 v d t$

$\Rightarrow x - 2 = 0 \int 3 2 t d t$

$\Rightarrow x - 2 = [\frac{2 t ^{2}}{2}]_{0}^{3}$

$\Rightarrow x = 11 m$