Ship A is sailing towards north -east with ve | vedclass

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

Question

$\begin{array}{l}
if\,we\,take\,the\,position\,of\,ship\,'A'\,as\,\
origin\,then\,positions\,velocity\,of\,both\
ships\,can\,be\,given\,a

Vedclass · Accepted Answer

$2.6$

Vedclass · Answer

$\begin{array}{l}
if\,we\,take\,the\,position\,of\,ship\,'A'\,as\,\
origin\,then\,positions\,velocity\,of\,both\
ships\,can\,be\,given\,as:\
{{\vec v}_A} = \left( {30\hat i + 50\hat j} ight)km/hr\
{{\vec v}_B} =  - 10\hat i\,km/hr\,\,;\,{{\vec r}_A} = 0\hat i + \hat j\
{{\vec r}_B} = \left( {80\hat i + 150\hat j} ight)\,km\
Time\,after\,which\,{m{distance}}\,between\
them\,will\,be\,\min imum
\end{array}$

$\begin{array}{l}
t = \frac{{{{\vec r}_{BA}} \cdot {{\vec v}_{BA}}}}{{\left| {{{\vec v}_{BA}}} ight|}}\,;\
Where\,{{\vec r}_{BA}} = \left( {80\hat i + 150\hat j} ight)\,km\
{{\vec v}_{BA}} =  - 10\left( {30\hat i + 50\hat j} ight)\,\left( { - 40\hat i - 50\hat j} ight)km/hr\
	herefore \,t =  - \frac{{\left( {80\hat i + 150\hat j} ight)\,.\left( { - 40\hat i - 50\hat h} ight)}}{{{{\left| { - 40i - 50\hat h} ight|}^2}}}\
\,\,\,\,\,\, = \,\frac{{3200 + 7500}}{{4100}}hr = \frac{{10700}}{{4100}}hr = 2.6\,hrs
\end{array}$

Similar Questions

The figure shows a velocity-time graph of a particle moving along a straight line The correct displacement-time graph of the particle is shown as

A particle has initial velocity $\left( {2\hat i + 3\hat j} \right)$ and acceleration $\left( {0.3\hat i + 0.2\hat j} \right)$. The magnitude of velocity after $10\, seconds$ will be

A man moves in an open field such that after moving $10 \,m$ on a straight line, he makes a sharp turn of $60^{\circ}$ to his left. The total displacement just at the start of $8^{\text {th }}$ turn is equal to ........$m$

A particle moves along an arc of a circle of radius $R$ . Its velocity depends on the distance covered as $v = a\sqrt s$ , where $a$ is a constant then the angle $\alpha $ between the vector of the total acceleration and the vector of velocity as a function of $s$ will be

A particle is moving with velocity $\vec v = K(y\hat i + x\hat j)$ where $K$ is a constant. The general equation for its path is