A police party is moving in a jeep at a const | vedclass

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(C) Let the police party catch the thief after time $t$.Distance traveled by the police party in time $t$ is $d_p = v t$.Distance traveled by the thie

Vedclass · Accepted Answer

$v^2 \geq 2 a x$

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(C) Let the police party catch the thief after time $t$.Distance traveled by the police party in time $t$ is $d_p = v t$.Distance traveled by the thief in time $t$ starting from rest with acceleration $a$ is $d_t = x + \frac{1}{2} a t^2$.For the police to catch the thief, the distance traveled by the police must be greater than or equal to the distance traveled by the thief:$v t \geq x + \frac{1}{2} a t^2$Rearranging the terms, we get a quadratic inequality in $t$:$\frac{1}{2} a t^2 - v t + x \leq 0$For a real solution for $t$ to exist, the discriminant of the corresponding quadratic equation $\frac{1}{2} a t^2 - v t + x = 0$ must be greater than or equal to zero.The discriminant $D = b^2 - 4ac$ is given by:$D = (-v)^2 - 4(\frac{1}{2} a)(x) \geq 0$$v^2 - 2 a x \geq 0$$v^2 \geq 2 a x$

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