Velocity of a particle moving along positive $x$ -direction is $v = (40 - 10 t)$ . Here, $t$ is in seconds At time $t = 0$ , the $x$ -coordinate of particle is zero. Find the time when the particle is at a point $(- 60 m, 0)$ from origin.

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$t_{3} = 3 (2 + \sqrt{7}) s$

$t_{3} = 2 (2 + \sqrt{8}) s$

$t_{3} = 4 (2 + \sqrt{7}) s$

$t_{3} = 2 (2 + \sqrt{7}) s$

answer is D.

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Detailed Solution

Comparing with $v = u + a t$ , we have $u = 40 m / s$ and $a = - 10 m / s^{2}$

Distance of $60 m$ from origin may be at $x = + 60 m$ and $x = - 60 m$ .
From the figure, we can see that at these two points particle is at three times, $t_{1}, t_{2}$ and $t_{3}$ .
For $X = + 60 m$ or $t_{1}$ and $t_{2}$
$s = u t + \frac{1}{2} a t^{2} \Rightarrow 60 = (+ 40) t + \frac{1}{2} (- 10) t^{2}$
Solving this equation, we get
$t_{1} = 2 s and t_{2} = 6 s$

For $X = - 60$ mor $t_{1}$
$s = u i + \frac{1}{2} a t^{2} - 60 = (+ 40) t + \frac{1}{2} (- 10) t^{2}$
Solving this equation, we get positive value of $t$ is $t_{3} = 2 (2 + \sqrt{7}) s$