(A) Let the position of the first cyclist be $A$ and the second cyclist be $B$ at time $t$. Let the distance between them be $x$.Given $OA = 4t$ and $

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) Let the position of the first cyclist be $A$ and the second cyclist be $B$ at time $t$. Let the distance between them be $x$.Given $OA = 4t$ and $OB = 3t$.Using the Law of Cosines in $\triangle OAB$:$x^2 = (OA)^2 + (OB)^2 - 2(OA)(OB) \cos(120^{\circ})$$x^2 = (4t)^2 + (3t)^2 - 2(4t)(3t) \left(-\frac{1}{2}\right)$$x^2 = 16t^2 + 9t^2 + 12t^2$$x^2 = 37t^2$$x = \sqrt{37}t$The rate at which they are separating is $\frac{dx}{dt}$.$\frac{dx}{dt} = \frac{d}{dt}(\sqrt{37}t) = \sqrt{37} \text{ km/h}$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

Difficult

Two cyclists are moving from a junction of two roads inclined at an angle of $120^{\circ}$ to each other,with velocities of $4 \text{ km/h}$ and $3 \text{ km/h}$ respectively. The rate at which they are separating from each other after $1 \text{ hour}$ is ..... $\text{km/h}$.

A
$\sqrt{37}$
B
$37$
C
$13$
D
$\sqrt{13}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Applications of Derivatives

Subtopic

Rate of Change of Quantities

$A$ spherical iron ball of $10 \ cm$ radius is coated with a layer of ice of uniform thickness that melts at a rate of $50 \ cm^3/min$. When the thickness of ice is $5 \ cm$,then the rate (in $cm/min$) at which the thickness of ice decreases is:

MediumAP EAMCET 2021

View Solution

$A$ particle moves such that $S = 6 + 48t - t^3$. The direction of motion reverses after moving a distance of

Medium

View Solution

If the radius $r$ of a sphere is increasing at a rate of $2 \text{ cm/s}$,then the rate of change of its surface area is proportional to which of the following?

Easy

View Solution

$A$ spherical balloon is expanding. If at any instant the rate of increase of its volume is $16$ times the rate of increase of its radius,then its radius at that instant is:

View Solution

The relation between pressure $p$ and volume $V$ is given by $p V^{1/4} = \text{constant}$. If the percentage decrease in volume is $\frac{1}{2} \%$, then the percentage increase in pressure is:

DifficultTS EAMCET 2013

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Two cyclists are moving from a junction of two roads inclined at an angle of $120^{\circ}$ to each other,with velocities of $4 \text{ km/h}$ and $3 \text{ km/h}$ respectively. The rate at which they are separating from each other after $1 \text{ hour}$ is ..... $\text{km/h}$.

Similar Questions

$A$ spherical iron ball of $10 \ cm$ radius is coated with a layer of ice of uniform thickness that melts at a rate of $50 \ cm^3/min$. When the thickness of ice is $5 \ cm$,then the rate (in $cm/min$) at which the thickness of ice decreases is:

$A$ particle moves such that $S = 6 + 48t - t^3$. The direction of motion reverses after moving a distance of

If the radius $r$ of a sphere is increasing at a rate of $2 \text{ cm/s}$,then the rate of change of its surface area is proportional to which of the following?

$A$ spherical balloon is expanding. If at any instant the rate of increase of its volume is $16$ times the rate of increase of its radius,then its radius at that instant is:

The relation between pressure $p$ and volume $V$ is given by $p V^{1/4} = \text{constant}$. If the percentage decrease in volume is $\frac{1}{2} \%$, then the percentage increase in pressure is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module