A jeep chases a thief with a constant velocity $v$ $v$ . When the jeep is at distance d from the thief, he starts to run with a constant acceleration $a$ $a$ . Show that the police will be able to catch the thief when $v \geq \sqrt{2 a d}$ $v >= sqrt(2ad)$ ?

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A jeep chases a thief with a constant velocity $v$ $v$ . When the jeep is at distance d from the thief, he starts to run with a constant acceleration $a$ $a$ . Show that the police will be able to catch the thief when $v \geq \sqrt{2 a d}$ $v >= sqrt(2ad)$ ?

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Answer 1 · 2017-05-22T09:17:59

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Explanation:

Given is speed of police jeep $= v$ $= v$

Distance between the thief and the police jeep $= d$ $=d$

At this point constant acceleration of thief $= a$ $= a$ .

Let the time when both meet $= t$ $=t$
Let $s$ $s$ be the distance covered by thief.
Using kinematic expression
$s = u t + \frac{1}{2} a t^{2}$ $s = ut+1/2 at^2$
we get
$s = \frac{1}{2} a t^{2}$ $s = 1/2 at^2$ ......(1)
During this time interval distance traveled by police jeep $= s + d$ $= s + d$
Which is also $= v t$ $= vt$

Equating two we get
$s + d = v t$ $s+d=vt$
$\Rightarrow s = v t - d$ $=>s = vt - d$ ......(2)
From (1) and (2) we get
$\frac{1}{2} a t^{2} = v t - d$ $1/2 at^2=vt - d$
$\Rightarrow a t^{2} - 2 v t + 2 d = 0$ $=>at^2 - 2vt + 2d = 0$

Solving quadratic for $t$ $t$
$t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $t=(-b+-sqrt(b^2-4ac))/(2a)$
$t = \frac{2 v \pm \sqrt{4 v^{2} - 4 a \times 2 d}}{2 a}$ $t=(2v+-sqrt(4v^2-4axx2d))/(2a)$
$\Rightarrow t = \frac{v \pm \sqrt{v^{2} - 2 a d}}{a}$ $=>t=(v+-sqrt(v^2-2ad))/(a)$
For a real time $t$ $t$ , the discriminant must be $\geq 0$ $>=0$ and $t$ $t$ can not be negative. From the first condition we get
$v^{2} - 2 a d \geq 0$ $v^2-2ad>=0$
$\Rightarrow v^{2} \geq 2 a d$ $=>v^2>=2ad$
$\Rightarrow v \geq \sqrt{2 a d}$ $=>v>=sqrt(2ad)$

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A jeep chases a thief with a constant velocity $v$ $v$ . When the jeep is at distance d from the thief, he starts to run with a constant acceleration $a$ $a$ . Show that the police will be able to catch the thief when $v \geq \sqrt{2 a d}$ $v >= sqrt(2ad)$ ?

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Explanation:

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A jeep chases a thief with a constant velocity vv. When the jeep is at distance d from the thief, he starts to run with a constant acceleration aa. Show that the police will be able to catch the thief when v≥√2adv >= sqrt(2ad)?

1 Answer

Explanation:

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A jeep chases a thief with a constant velocity $v$ $v$ . When the jeep is at distance d from the thief, he starts to run with a constant acceleration $a$ $a$ . Show that the police will be able to catch the thief when $v \geq \sqrt{2 a d}$ $v >= sqrt(2ad)$ ?