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 car 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???
plz answer​

Answer 1

Let both the cars A and B meet at C which is at $\sf{x}$ kilometres away from initial position of car B, after time $\sf{t}$ seconds.

The car B moves with speed of $\sf{10\ km\,s^{-1}}$ without acceleration.

By second equation of motion, displacement of car B will be,

$\sf{\longrightarrow x=10t\quad\quad\dots (1)}$

The car A moves with speed of $\sf{10\ km\,s^{-1}}$ but has an acceleration of $\sf{4\ km\,s^{-2}.}$

By second equation of motion, displacement of car A will be,

$\sf{\longrightarrow 0.5+x=10t+\dfrac{1}{2}(4)t^2}$

$\sf{\longrightarrow 0.5+x=10t+2t^2\quad\quad\dots(2)}$

Subtracting (1) from (2),

$\sf{\longrightarrow 0.5=2t^2 }$

$\sf{\longrightarrow t=0.5\ s }$

Then (1) becomes,

$\sf{\longrightarrow\underline{\underline{x=5\ km }}}$

Hence 5 km is the answer.

Answer 2

Let both the cars A and B meet at C which is at x kilometres away from initial position of car B, after time t seconds.

The car B moves with speed of

${10kms}^{ - 1}$

without acceleration.

By second equation of motion, displacement of car B will be,

⟶ x = 10t…(1)

The car A moves with speed of ${10kms}^{ - 1}$ but has an acceleration of ${4kms}^{ - 2}$ .

By second equation of motion, displacement of car A will be,

$⟶0.5+x=10t+21(4)t2\sf{\longrightarrow0.5+x=10t+2t^2\quad\quad\dots(2)}$

Subtracting (1) from (2),

\sf{\longrightarrow 0.5=2t^2 }⟶0.5=2t

2

\sf{\longrightarrow t=0.5\ s }⟶t=0.5 s

Then (1) becomes,

\sf{\longrightarrow\underline{\underline{x=5\ km }}}⟶

x=5 km

Hence 5 km is the answer.