You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Inequalities Involving Modulus Function

Two cars star...

Question

Two cars started from a place one moving due east and other due north with equal speed V. Then the rate at which they were being separated from from each other is

\frac{\sqrt{2}}{V}

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

\frac{V}{\sqrt{2}}

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

\frac{1}{\sqrt{2} V}

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

\sqrt{2 V}

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

Open in App

Solution

The correct option is A $\frac{\sqrt{2}}{V}$
After time t distance travelled by $1^{s t}$ car = vt
second car = vt
Rate at which they are separated = $\sqrt{2} v t$
Rate of speed = $\sqrt{2} v$

Suggest Corrections

Similar questions

Q. Car

A

is moving due east with speed

v_{1}

. At a certain time it is

l_{1}

away from the crossing.Car

B

at that instant is

l_{2}

away from the crossing and moving due north with a speed

v_{2}

. Find the shortest distance between them.

Two men $P$ , and $Q$ start with velocities $v$ at the same time from the junction of two roads inclined at $45^{\circ}$ to each other. If they travel by different roads, the rate at which they are being separated, is

Two men A and B start with velocities V at the same time from the junction of two roads inclined at $45^{\circ}$ to each other. If they travel by different roads, then find the rate at which they are being separated.

Q. A car is moving towards north with a speed of

5 m / s

. A motorcyclist is moving towards east with the speed of

5 m / s

. At time

t = 0

, the motorcyclist is in due north of the car and is at a distance of

30 m

from the car. The time

t

at which they will be closest to each other is

Two identical particles move towards each other with velocity 2v and v respectively. The velocity of the centre of mass is

Join BYJU'S Learning Program

Two cars started from a place one moving due east and other due north with equal speed V. Then the rate at which they were being separated from from each other is

The correct option is A √2VAfter time t distance travelled by 1st car = vtsecond car = vtRate at which they are separated = √2vtRate of speed = √2v

The correct option is A $\frac{\sqrt{2}}{V}$
After time t distance travelled by $1^{s t}$ car = vt
second car = vt
Rate at which they are separated = $\sqrt{2} v t$
Rate of speed = $\sqrt{2} v$