Physics, asked by sauravjadhav2468, 4 months ago

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​

Answers

Answered by shadowsabers03
108

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.

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Answered by ItzIshan
77

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

Given :-

  • Car a and b is moving at the same speed of 10 km/s before acceleration.

  • Car b is ahead of car a

  • Distance between car a and b is 500m

  • Car a is accelerated by 4 km/

To FinD :-

  • At what Distance both car will meet.

AnsweR :-

Car a and car b are 500 m apart =

 \sf \:  =  \frac{500}{1000} km \\  \\  \sf \:  = 0.5 \: km

Let us take C is the point where Car a and b will meet and 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

We know that :-

second equation of motion

  •  \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/

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

Distance =(0.5 + x) km

Time = t

Acceleration = 4 km/

 \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)

For the car b ,

Distance = x

Time = t

We know that

  •  \sf \: distance = time \times speed

 \sf  \implies\: x = 10t -  -  -  - (ii)

From the Equation (i) and (ii) :-

\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}

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

\implies \sf \:  \: x = 10t \\  \\ \implies \sf \:  \: x = 10 \times 0.5 \\  \\  \implies \boxed{ \sf \:  \: x = 5 \: km \: }

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

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Hope it will help you :)

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