Question

Two cars aand b are at rest at the origin o. if a starts with uniform velocity 20m/s at thew same direction with constan​

Answer 1

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=>Let distance covered by A be S1

=>Let distance covered by B be S2

B will catch A when S1 = S2

B will catch A when S1 = S2S = ut + 1/2 × a × t²

Therefore,

40t + (1÷2 × 0 × t²) = 0t + 1÷2 × 4 × t²

40t + (1÷2 × 0 × t²) = 0t + 1÷2 × 4 × t²40t = 2t²

40t + (1÷2 × 0 × t²) = 0t + 1÷2 × 4 × t²40t = 2t²20t = t²

Therefore, t = 20

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Answer 2

$\huge \mathfrak \pink {answer}$

Cars A and B start from origin. Car A travels with a uniform velocity of 20 m/s. Car B starts with a constant acceleration of 2 m/s².

To find:

Time after which they will meet.

Concept:

Both the cars will meet only after covering an equal distance. So , we need to find out the distance and equate them.

Calculation:

Distance travelled by car A in time t :

$d1 = v \times t$

$= > d1 = 20 \times t$

$= > d1 = 20t$

Distance travelled by car B in time t :

$d2 = ut + \frac{1}{2} a {t}^{2}$

$= > d2 = 0 + \{ \frac{1}{2} \times 2 \times {t}^{2} \}$

$= > d2 = {t}^{2}$

Equating the 2 distances travelled :

$</p><p>d1 = d2$

$= > 20t = {t}^{2}$

$</p><p> = > {t}^{2} - 20t = 0$

$= > t(t - 20) = 0$

So , w can say that :

$</p><p> \therefore \: \: t - 20 = 0$

$= > t = 20 \: seconds$

So they will meet after 20 secs.

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