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

$\huge\sf{\underline{ \bold{Q}uestion}}:$

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 with the same speed of 10 km/s. If vehicle a is accelerated by 4 km/s² and car b is having same speed as earlier, find at what distance car a and b will meet.

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$\huge\sf{\underline{To \: find}} \: –$

At what distance both car a and b will meet ?

$\large\bf{\underline{Given}} \: –$

• Car b is ahead of car a.
• Distance between car a and b is 500m.
• Car a and b is moving at the same speed of 10 km/s before acceleration.
• Car a is accelerated by 4 km/s².

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$\huge\bold{\underline{\underline{Solution}}}:$

Car a and car b are 500 m apart =

$\begin{gathered} \sf \: = \frac{500}{1000} km \\ \\ \sf \: = 0.5 \: km\end{gathered}$

Let us take C as the point where car a and b would meet.

Also, let us take C is x km away from initial position of car b after time t.

$\star \sf \: total \: distance \: covered \: by \: a \: and \: b = (0.5 + x)$⠀

• Speed of car b = 10 km/s

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We know that -

second equation of motion is :

• $\sf \: s = ut + \frac{1}{2} a {t}^{2}$

The car a is also moving with the same speed of 10 km/s. And it is accelerated by 4 km/s².

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According to the second equation of motion , the displacement of car a →

• Distance =(0.5 + x) km

• Time = t

• Acceleration = 4 km/s²

$\begin{gathered} \sf \: s = ut + \frac{1}{2} a {t}^{2} \\ \\ \implies \sf \: 0.5 + x = 10t + \frac{1}{2} \times 4 \: \times {t}^{2} \\ \\ \implies \sf \: 0.5 + x = 2 {t}^{2} - - - - (i)\end{gathered}$

For the car b,

• Distance = x

• Time = t

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We also know that,

$\sf \: distance = time \times speed \\ \sf \implies\: x = 10t - - - - (ii)$

From the Equation (i) and (ii), we get-

$\begin{gathered}\implies \sf \: 0.5 + x = 10t + 2 {t}^{2} \\ \\ \implies \sf \: \: 0.5 + x = 10 \times \frac{x}{10} + 2 {t}^{2} \\ \\ \implies \sf \: \: 0.5 + x = x + 2 {t}^{2} \\ \\ \implies \sf \: \: 2 {t}^{2} = 0.5 + x - x \\ \\ \implies \sf \: \: {t}^{2} = \frac{0.5}{2} \\ \\ \implies \sf \: \: {t}^{2} = 0.25 \\ \\ \implies \sf \: t= \sqrt{0.25} \\ \\ \implies \boxed{ \sf \: t = 0.5 \: s}\end{gathered}$

Substituting the value of t into Equation (ii) :-

$\begin{gathered}\implies \sf \: \: x = 10t \\ \\ \implies \sf \: \: x = 10 \times 0.5 \\ \\ \implies \boxed{ \sf \: \: x = 5 \: km \: }\end{gathered}$

Hence , The car a and b will meet after 5 kilometres.

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⠀⠀⠀✨HOPE IT HELPS✨

Answer 2

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.