Given: Velocity of ship A, v A = ( 30 i ̂ + 50 j ̂ ) k m / h , v_{A} = (30\widehat{i} + 50\widehat{j})km/h\text{,~} Velocity of ship B v ⃗ B = − 10 i ̂ k m / h {\overrightarrow{v}}_{B} = - 10\widehat{i}\text{\ }km/h Distance between two ships = r ⃗ B − r ⃗ A = 80 i ̂ + 150 j ̂ . = {\overrightarrow{r}}_{B} - {\overrightarrow{r}}_{A} = 80\widehat{i} + 150\widehat{j}\text{.~} To find: The time at which the distance between the two ships is minimum.At any time t t , the position of s h i p A shipA , x ⃗ A = ( 30 i ̂ + 50 j ̂ ) t {\overrightarrow{x}}_{A} = (30\widehat{i} + 50\widehat{j})t The position of ship B , B, x ⃗ B = ( − 10 t + 80 ) i ̂ + 150 j ̂ {\overrightarrow{x}}_{B} = ( - 10t + 80)\widehat{i} + 150\widehat{j} Distance between ship A A and B , B, x ⃗ A B = x ⃗ B − x ⃗ A {\overrightarrow{x}}_{AB} = {\overrightarrow{x}}_{B} - {\overrightarrow{x}}_{A} | x ⃗ A B | = ( − 10 t + 80 − 30 t ) 2 + ( 150 − 50 t ) 2 \left| {\overrightarrow{x}}_{AB} \right| = \sqrt{( - 10t + 80 - 30t)^{2} + (150 - 50t)^{2}} For | x ⃗ A B | \left| {\overrightarrow{x}}_{AB} \right| to be minimum, d d t x A B = 0 \frac{d}{dt}x_{AB} = 0 d d t ( 16 t 2 + 64 − 64 t + 225 + 25 t 2 − 150 t ) = 0 \frac{d}{dt}\left( 16t^{2} + 64 - 64t + 225 + 25t^{2} - 150t \right) = 0 32 t − 64 + 50 t − 150 = 0 32t - 64 + 50t - 150 = 0 t = 214 82 = 2.6 h t = \frac{214}{82} = 2.6\text{\ }h

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Question 1:

Ship A is sailing towards north-east with velocity $\overrightarrow{v} = \left( 30\widehat{i} + 50\widehat{j} \right)\ kmh^{- 1}$ where î points east and ĵ, north. Ship B is at a distance of 80 km east and 150 km north of ship A and is sailing towards west at 10 km h^-1. A will be at minimum distance from B in

$2.2\text{\ }h$

$3.2\text{\ }h$

$4.2\text{\ }h$

$2.6\text{\ }h$

Answer:

Solution:

Given: Velocity of ship A,

$v_{A} = (30\widehat{i} + 50\widehat{j})km/h\text{,~}$

Velocity of ship B

${\overrightarrow{v}}_{B} = - 10\widehat{i}\text{\ }km/h$

Distance between two ships

$= {\overrightarrow{r}}_{B} - {\overrightarrow{r}}_{A} = 80\widehat{i} + 150\widehat{j}\text{.~}$

To find: The time at which the distance between the two ships is minimum.

At any time $t$ , the position of $shipA$ ,

${\overrightarrow{x}}_{A} = (30\widehat{i} + 50\widehat{j})t$

The position of ship $B,$

${\overrightarrow{x}}_{B} = ( - 10t + 80)\widehat{i} + 150\widehat{j}$

Distance between ship $A$ and $B,$

${\overrightarrow{x}}_{AB} = {\overrightarrow{x}}_{B} - {\overrightarrow{x}}_{A}$

$\left| {\overrightarrow{x}}_{AB} \right| = \sqrt{( - 10t + 80 - 30t)^{2} + (150 - 50t)^{2}}$

For $\left| {\overrightarrow{x}}_{AB} \right|$ to be minimum,

$\frac{d}{dt}x_{AB} = 0$

$\frac{d}{dt}\left( 16t^{2} + 64 - 64t + 225 + 25t^{2} - 150t \right) = 0$

$32t - 64 + 50t - 150 = 0$

$t = \frac{214}{82} = 2.6\text{\ }h$

Prev Exam(s):

JEE MAIN

- 2019

, 8 Apr Shift-01

Chapter:

RELATIVE MOTION

Question Level:

Tags:

PHYSICS
RELATIVE MOTION