The speed of a body moving with uniform acceleration is $u$ . This speed is doubled while covering a distance $S$ . When it covers an additional distance $S$ , its speed would become ?

\sqrt{3} u

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

\sqrt{5} u

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

\sqrt{11} u

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

\sqrt{7} u

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

Open in App

Solution

The correct option is D $\sqrt{7} u$
Formula:

$v^{2} = u^{2} + 2 a s$

given,

$(2 u)^{2} = u^{2} + 2 a s$

$4 u^{2} = u^{2} + 2 a s$

$2 a s = 3 u^{2}$

$v^{2} = u^{2} + 2 a s$

$v^{2} = u^{2} + 2 (2 a s)$

$v^{2} = u^{2} + 2 (3 u^{2})$

$= 7 u^{2}$

$∴ v = \sqrt{7} u$

Suggest Corrections

Similar questions

A body starts moving with an initial velocity 50 m/s and acceleration 20 $m / s^{2}$ . How much distance it will cover in 4s ? Also, calculate its average speed during this time interval.

Q. The acceleration

α

of a particle moving with an velocity

u

varies with distance

x

a = k \sqrt{x}

where

k

is a constant . Th distance covered by the particle when its velocity become

3 u

is given by

A body covers a distance S in time t. What is its speed?

A particle moving in a straight line covers half the distance with a speed of 3m/s. The other half of distance is covered with double speed 6m/s. The average sped of the particle is:

A particle covers $50 m$ distance when projected with an initial speed. On the same surface, when projected with double the initial speed it will cover a distance of :

Join BYJU'S Learning Program

The speed of a body moving with uniform acceleration is u. This speed is doubled while covering a distance S. When it covers an additional distance S, its speed would become ?

The correct option is D √7uFormula:v2=u2+2asgiven,(2u)2=u2+2as4u2=u2+2as2as=3u2v2=u2+2asv2=u2+2(2as)v2=u2+2(3u2)=7u2∴v=√7u

The speed of a body moving with uniform acceleration is $u$ . This speed is doubled while covering a distance $S$ . When it covers an additional distance $S$ , its speed would become ?

The correct option is D $\sqrt{7} u$
Formula:

$v^{2} = u^{2} + 2 a s$

given,

$(2 u)^{2} = u^{2} + 2 a s$

$4 u^{2} = u^{2} + 2 a s$

$2 a s = 3 u^{2}$

$v^{2} = u^{2} + 2 a s$

$v^{2} = u^{2} + 2 (2 a s)$

$v^{2} = u^{2} + 2 (3 u^{2})$

$= 7 u^{2}$

$∴ v = \sqrt{7} u$