Question

two cars a and b are 500m apart from each other.car b is ahead of car a. they are moving in the same direction and the same peed of 10km/s if veachle a is accelerated by 4km/s2 and car b is having same speed as earlier find at what distance car a and b will meet ​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese} \:\:Given :- }}}$

$\sf\implies Two \ cars \ A\ \& \ B \ are \ 500m \ apart. \\\\\sf\implies They \ have \ intial \ Velocity\ of \ 10m/s. \\\\\sf\implies Car \ B \ is \ ahead \ of \ A. \\\\\sf\implies Car \ A \ has \ a \ acceleration \ of \ 4m / s^2.$

$\Large\underline{\underline{\red{\sf \purple{\maltese} \:\:To \ Find :- }}}$

$\sf\implies Distance\ after \ which \ both \ will \ meet.$

$\Large\underline{\underline{\red{\sf \purple{\maltese} \:\:Solution :- }}}$

Let us take the two cars will meet after travelling x distance. And let us take that it took a time of t for both the cars to meet . The initial Velocity of both cars were same . And the car A accelerated with the acceleration of 4m / s² . So ,

$\sf\implies Total \ distance \ travelled \ by \ A \ \& B \ = \ ( 500 + x ) m$

$\underline{\pink{\bf For \ Car \ A \ :- }}$

$\sf \implies Distance = ( 500 + x ) m \\\\\sf\implies Time = t \ seconds. \\\\\sf\implies Acceleration \ of \ the \ car = 4m/s^2 .$

$\orange{\boldsymbol{ Using \ second\ equ^n \ of \ motion :- }}$

$\tt:\implies \green{ s = ut + \dfrac{1}{2} at^2}\\\\\tt:\implies 500 + x = 10t + \dfrac{1}{2}\times 4m/s^2 \times t \\\\\tt:\implies \boxed{\red {\tt 500 + x = 10t + 2t^2 }}$

$\rule{200}2$

$\underline{\pink{\bf For \ Car \ B \ :- }}$

$\sf \implies Distance = x \\\\\sf\implies Time taken = t \ seconds.$

• So now ,

$\tt:\implies \green{ Distance = Speed \times Time }\\\\\tt:\implies x = 10m/s \times t \\\\\tt:\implies \boxed{\red {\tt x = 10t }}$

$\rule{200}2$

$\underline{\blue{\sf From\ above \ two \ equations :- }}$

$\tt:\implies 500 + x = 10t + 2t^2 \\\\\tt:\implies 500 + x = 10 \times \dfrac{x}{10} + 2t^2\\\\\tt:\implies 500+x = x + 2t^2 \\\\\tt:\implies 2t^2 = 500 \\\\\tt:\implies t^2 =\dfrac{500}{2}\\\\\tt:\implies t^2 = 250 \\\\\tt:\implies t = \sqrt{25 \times 10 }\\\\\underline{\boxed{\red{\tt\longmapsto Time= 5\sqrt{10}s }}}$

$\rule{200}2$

$\underline{\blue{\sf Put \ t = 5\sqrt{10} \ in \ (ii) :- }}$

$\tt:\implies x = 10 t \\\\\tt:\implies x = 10 \times 5 \sqrt{10} \\\\\underline{\boxed{\red{\tt\longmapsto Distance= 50\sqrt{10}m}}}$

Hence they will meet after travelling 50√10 m.