A point particle starting from rest has a velocity that increases linearly with time such that v = pt, Where p = 4 $m s^{- 2}$ . The distance covered in the first 2 s Will be

6 m

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

4 m

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

8 m

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

10 m

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

Open in App

Solution

The correct option is C 8 m
Given that,

Velocity $v = p t$

$P = 4$ m/s²

$t = 2$ sec

Now,

$s = u t + \frac{1}{2} a t^{2}$

$s = 0 + \frac{1}{2} \times 4 \times 4$

$s = 8 m$

Hence, the distance covered in the first 2 sec is 8 m.

Suggest Corrections

Similar questions

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

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

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

A point particle starting from rest has a velocity that increases linearly with time such that v = pt, Where p = 4 ms−2. The distance covered in the first 2 s Will be

The correct option is C 8 mGiven that, Velocity v=pt P=4 m/s2 t=2 sec Now, s=ut+12at2 s=0+12×4×4 s=8m Hence, the distance covered in the first 2 sec is 8 m.

A point particle starting from rest has a velocity that increases linearly with time such that v = pt, Where p = 4 $m s^{- 2}$ . The distance covered in the first 2 s Will be

The correct option is C 8 m
Given that,

Velocity $v = p t$

$P = 4$ m/s²

$t = 2$ sec

Now,

$s = u t + \frac{1}{2} a t^{2}$

$s = 0 + \frac{1}{2} \times 4 \times 4$

$s = 8 m$

Hence, the distance covered in the first 2 sec is 8 m.