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Standard IX

Physics

Derivation of Position-Velocity Relation by Graphical Method

Initially car...

Question

Initially car A is 21 m ahead of car B. Both start moving at time t=0 in same direction along straight line. if A is moving with constant velocity 20 m/s and B starts from rest with an acceleration ${2 m / s}^{2}$ towards A, then the time when car B will catch the car A is

21 s

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22 s

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10 s

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C a n n e v e r c a t c h A

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Solution

The correct option is A $21 s$
The relative speed of $B$ wrt $A$ will be $V_{B A} = V_{B} - V_{A} = 20 - 0 = - 20 m / s$
their relative acceleration will be $a_{B A} = a_{B} - a_{A} = 2 - 0 = 2 m / s^{2}$
the distance traveled will be given by $s = u t + \frac{1}{2} a t^{2} = - 20 t + \frac{1}{2} 2 \times (t)^{2} = 21 m e t e r$
on solving it we get $t = 21 s e c$
Just employ the Shridharacharya Formula $t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Where $t$ is root of $a t^{2} + b t + c = 0$

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Initially car A is 21 m ahead of car B. Both start moving at time t=0 in same direction along straight line. if A is moving with constant velocity 20 m/s and B starts from rest with an acceleration 2m/s2 towards A, then the time when car B will catch the car A is

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