Question

two cars A and B are moving in same direction with velocity 30 metre per second in 20 metre per second when car a is at a distance D behind the kababi the driver of the car in applies brakes producing uniform retardation of 2 metre per second square there will be no collagen when
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Answer 1

$\sf{\underline{Answer:}}$ d is greater than 25 m.

$\sf{\underline{Therefore:}}$

The correct option is: (3) d > 25 m

$\sf{\underline{Solution:}}$

For no collision the velocity of A must reduce to that of velocity of B.

$\sf{\underline{Here:}}$

$\implies$ $\sf{XA = 30 \: m/s}$

$\implies$ $\sf{XB = 20 \: m/s}$

$\implies$ $\sf{a = - 2 \: {m/s}^{2}}$

$\sf{\underline{That\:is:}}$

$\sf{\underline{Relative\:final\:velocity\:of:}}$

(i) A with respect to B is: $\boxed{\sf{0. Xr = 0}}$

$\sf{\underline{Relative\:initial\:velocity\:of:}}$

(ii) A with respect to B is: $\boxed{\sf{( XA-XB )}}$

$\sf{\underline{Acceleration\:of:}}$

(iii) A with respect to B is: $\boxed{\sf{(-a - 0 ) = - a}}$

$\sf{\underline{Using\:the\:equation:}}$

$\boxed{\sf{ {v}^{2} = {u}^{2} + 2as}}$

$\implies$ $\sf{0 = (XA - XB)2 + 2 \times - a \times s}$

$\implies$ $\sf{s = \frac{ {(XA - XB)}^{2} }{2a}}$

$\implies$ $\sf{s = \frac{ ({30 - 20})^{2} }{4}}$

$\implies$ $\sf{s = \frac{( {10})^{2} }{4}}$

$\implies$ $\sf{s = \frac{100}{4}}$

$\implies$ $\sf{s = 25}$

$\sf{\underline{Therefore:}}$

No collision occurs when d is greater than 25 m.

Answer 2

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