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Standard IX

Physics

Derivation of Position-Velocity Relation by Graphical Method

Car A and car...

Question

Car A and car B start moving simultaneously in the same direction along the line joining them car A with a constant acceleration $a = 4 m / s^{2}$ while car B moves with a constant velocity V = 1 m/s. At t = 0 car A is 10 m behind car B. Find the time when car A over takes car B.

2 s

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2.5 s

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1 s

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1.5 s

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Solution

The correct option is B 2.5 s
The relative speed of $A$ wrt $B$ will be $V_{A B} = V_{A} - V_{B} = 0 - 1 = - 1 m / s$
their relative acceleration will be $a_{A B} = a_{A} - a_{B} = 4 - 0 = 4 m / s^{2}$
the distance traveled will be given by $s = u t + \frac{1}{2} a t^{2} = (- 1) t + \frac{1}{2} 4 \times (t)^{2} = 10 m e t e r$
or $2 t^{2} - t - 10 = 0$
on solving it we get $t = 2.5 s e c$
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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Car A and car B start moving simultaneously in the same direction along the line joining them car A with a constant acceleration a=4m/s2 while car B moves with a constant velocity V = 1 m/s. At t = 0 car A is 10 m behind car B. Find the time when car A over takes car B.

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