The velocity-time relation of an electron starting from rest is given by $u = k t$ , where $k = 2 m / s^{2}$ The distance traversed in $3$ sec is:-

9 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

16 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

27 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

36 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $9 m$

$u = k t$

$\frac{d x}{d t} = k t$
$\Rightarrow d x = k t d t$

or $\int_{0}^{x} d x = \int_{0}^{3} k t d t$

$x = k {(\frac{t^{2}}{2})}_{0}^{3} = k (\frac{9}{2}) = 2 (\frac{9}{2}) = 9 m$

Suggest Corrections

Similar questions

An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m s^{- 2}$ . The distance travelled in the first 3 seconds will be

Q. An electron starting from rest has a velocity that increases linearly with time, i.e.,

v = k t

, where

k = 2 m / s^{2}

. The distance covered in the first

3 s

Q. An electron starting from rest has a velocity that increases linearly with time that is

v = k t

where

k = 2 m / s^{2}

. Distance covered in first

3 s

will be:

Q. An electron starting from rest has a velocity that increases linearly with time,i.e. v=kt where

k = 2 m / s^{2}

. The distance covered in the first three seconds will be

An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m s^{- 2}$ . The distance travelled in the first 3 seconds will be

Join BYJU'S Learning Program

The velocity-time relation of an electron starting from rest is given by u=kt, where k=2 m/s2 The distance traversed in 3 sec is:-

The correct option is A 9m u=kt dxdt=kt ⇒dx=ktdt or∫x0dx=∫30ktdt x=k(t22)30=k(92)=2(92)=9m

The velocity-time relation of an electron starting from rest is given by $u = k t$ , where $k = 2 m / s^{2}$ The distance traversed in $3$ sec is:-

The correct option is A $9 m$

$u = k t$

$\frac{d x}{d t} = k t$
$\Rightarrow d x = k t d t$

or $\int_{0}^{x} d x = \int_{0}^{3} k t d t$

$x = k {(\frac{t^{2}}{2})}_{0}^{3} = k (\frac{9}{2}) = 2 (\frac{9}{2}) = 9 m$