Question

The distance between Kanpur & Jaipur is 90 km.A started from Jaipur with 5 kmph and B started from Kanpur with 3 kmph.B increases his speed 1.5every hour.when will they meet?

Answer 1

i hope it helped you

Answer 2

"Answer: They met after 17.94 hours.

Given,

Speed of A from Jaipur = 5 kmph = 1.388 m/s

Speed of B from Kanpur = 3 kmph = 0.833 m/s

Acceleration of B $= 1.5\quad \frac { m }{ { s }^{ 2 } }$

Distance between Jaipur and Kanpur = 90 km = 90000 m

Solution:

For the person A:

From the diagram, the distance of A = 90 - x

$Speed\quad =\quad \frac { Distance }{ Time }$

$1.388\quad =\quad \frac { 90000\quad -\quad x }{ t }$

$1.388t\quad =\quad 90000\quad -\quad x$

$x\quad =\quad 90000\quad -\quad 1.388t \quad \longrightarrow \quad \left( 1 \right)$

For the person B:

From the diagram, the distance of B = x

As the acceleration is given for B, then we can use the below given formula,

$s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } a{ t }^{ 2 }$

$x\quad =\quad 0.833t\quad +\quad 0.75{ t }^{ 2 } \quad \longrightarrow \quad \left( 2 \right)$

On equation (1) and (2), we get,

$90000\quad -\quad 1.388t \quad =\quad 0.833t\quad +\quad 0.75{ t }^{ 2 }$

$0.75{ t }^{ 2 }\quad +\quad 2.221t\quad -\quad 90000\quad =\quad 0$

To solve the above equation, we need to use,

$x\quad =\quad \frac { -b\quad \pm \quad \sqrt { { b }^{ 2 }\quad -\quad 4ac } }{ 2a }$

On solving, we get

$x\quad =\quad \frac { -b\quad \pm \quad \sqrt { { b }^{ 2 }\quad -\quad 4ac }}{ 2a }$

$a\quad =\quad 0.75;\quad b\quad =\quad 2.221;\quad c\quad =\quad -90000$

$x\quad =\quad \frac { -2.221\quad \pm \quad \sqrt { 4.932\quad -\quad \left( 4\quad \times \quad 0.75\quad \times \quad -90000\right)}}{2\quad \times \quad 0.75 }$

$x\quad =\quad \frac { -2.221\quad \pm \quad 519.61 }{ 1.5 }$

$x\quad =\quad \frac { -2.221\quad -\quad 519.61 }{ 1.5 } ,\quad \frac { -2.221\quad +\quad 519.61 }{ 1.5 }$

$x\quad =\quad -347.88,\quad 344.92$

On eliminating the negative value, we get,

x = 344.92

On substituting the value of x on equation (1), we get

$344.92\quad =\quad 90000\quad -\quad 1.388t$

t = 64592.99 seconds

t = 17.94 hours"