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Standard XIII

Physics

1st Equation of Motion

Car A and Car...

Question

Car A and Car B starts from origin with speeds $54$ kmph and $72$ kmph respectively. Both starts from origin at the same time. Find the distance between A and B after $20$ minutes .

54

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30

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24

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Solution

The correct option is B $30$ km
Given $t = 20 m i n = \frac{1}{3} h r$

$X_{A} = v_{1} t = 54 \times \frac{1}{3} = 18 k m$
$Y_{B} = v_{2} t = 72 \times \frac{1}{3} = 24 k m$
Distance between A and B is
$d = \sqrt{X_{A}^{2} + Y_{B}^{2}} = \sqrt{18^{2} + 24^{2}} = 6 \sqrt{3^{2} + 4^{2}} = 30 k m$

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Car A and Car B starts from origin with speeds 54 kmph and 72 kmph respectively. Both starts from origin at the same time. Find the distance between A and B after 20 minutes .

The correct option is B 30 kmGiven t=20 min=13hr XA=v1t=54×13=18 km YB=v2t=72×13=24 km Distance between A and B is d=√X2A+Y2B=√182+242=6√32+42=30 km

Car A and Car B starts from origin with speeds $54$ kmph and $72$ kmph respectively. Both starts from origin at the same time. Find the distance between A and B after $20$ minutes .

The correct option is B $30$ km
Given $t = 20 m i n = \frac{1}{3} h r$

$X_{A} = v_{1} t = 54 \times \frac{1}{3} = 18 k m$
$Y_{B} = v_{2} t = 72 \times \frac{1}{3} = 24 k m$
Distance between A and B is
$d = \sqrt{X_{A}^{2} + Y_{B}^{2}} = \sqrt{18^{2} + 24^{2}} = 6 \sqrt{3^{2} + 4^{2}} = 30 k m$