An electron starting from rest has a velocity v that increases linearly with time t as v=kt, where k=2ms-2 . The distance covered by electron in first 3 s is

Question

An electron starting from rest has a velocity v that increases linearly with time t as v=kt, where k=2ms-2 . The distance covered by electron in first 3 s is (A) 9m (B) 11 m (C) 33m (D) 40m

Tardigrade Admin · Accepted Answer

Given, v=kt Also k=2ms-2 ∴ v=2t or (dx/dt)=2t On integration of both sides x=t2+c When, t=0, then x=0 ∴ 0=(0)2+c ⇒ c=0 So, we get x=t2 When t=3 ⇒ x=(3)2=9m

Q. An electron starting from rest has a velocity $v$ that increases linearly with time $t$ as $v = k t,$ where $k = 2 m s^{- 2}$ . The distance covered by electron in first $3 s$ is

9m

11 m

33m

40m

Given, $v = k t$
Also $k = 2 m s^{- 2}$
$∴$ $v = 2 t$
or $\frac{d x}{d t} = 2 t$
On integration of both sides
$x = t^{2} + c$
When, $t = 0,$ then $x = 0$
$∴$ $0 = (0)^{2} + c$
$\Rightarrow$ $c = 0$
So, we get $x = t^{2}$
When $t = 3$
$\Rightarrow$ $x = (3)^{2} = 9 m$

Q. An electron starting from rest has a velocity v that increases linearly with time t as v=kt, where k=2ms−2 . The distance covered by electron in first 3s is

9m

11 m

33m

40m

Given, v=kt Also k=2ms−2 ∴ v=2t or dtdx​=2t On integration of both sides x=t2+c When, t=0, then x=0 ∴ 0=(0)2+c ⇒ c=0 So, we get x=t2 When t=3 ⇒ x=(3)2=9m

Q. An electron starting from rest has a velocity $v$ that increases linearly with time $t$ as $v = k t,$ where $k = 2 m s^{- 2}$ . The distance covered by electron in first $3 s$ is

Given, $v = k t$
Also $k = 2 m s^{- 2}$
$∴$ $v = 2 t$
or $\frac{d x}{d t} = 2 t$
On integration of both sides
$x = t^{2} + c$
When, $t = 0,$ then $x = 0$
$∴$ $0 = (0)^{2} + c$
$\Rightarrow$ $c = 0$
So, we get $x = t^{2}$
When $t = 3$
$\Rightarrow$ $x = (3)^{2} = 9 m$