Question

At the moment $t = 0$ a particle leaves the origin and moves in the positive direction of the $x$ -axis. Its velocity varies with time as $v = v_{0} (1 - \frac{t}{τ})$ , where $v_{0}$ is the initial velocity vector whose modulus equals $v_{0} = 10.0 cm s^{- 1}, τ = 5.0 s$ . Find,
$(a)$ the $x$ coordinate of the particle at the moments of time $6.0, 10$ and $20 s$ ,
$(b)$ the moments of time when the particle is at the distance $10.0 cm$ from the origin,

$(c)$ the distance $s$ covered by the particle during the first $4.0 s$ and $8.0 s$ ; draw the approximate plot $s (t)$ .

Accepted Answer

Given, at time $t = 0$ , position $x = 0$ and velocity $v = v_{0} (1 - \frac{t}{τ}) . . . (1)$ .

Let at time $t = t$ position be $x = x$ .

$(a)$ As velocity is given by $v = \frac{d x}{d t}$ , so we have,

$d x = v d t \Rightarrow d x = v_{0} (1 - \frac{t}{τ}) d t$

Integrating within proper limits, we get,
$\int_{0}^{x} x = \int_{0}^{t} v_{0} (1 - \frac{t}{τ}) d t \Rightarrow x = v_{0} (t - \frac{t^{2}}{2 τ}) \Rightarrow x = v_{0} t (1 - \frac{t}{2 τ}) . . . (2)$
Hence, we can find $x$ coordinates of the particle at $t = 6.0 s, 10 s$ and $20 s$ .

At $t = 6 s$ ,
$x = 10 \times 6 (1 - \frac{6}{2 \times 5}) = 24 cm = 0.24 m$

At $t = 10 s$ ,
$x = 10 \times 10 (1 - \frac{10}{2 \times 5}) = 0$
At $t = 20 s$ ,

$x = 10 \times 20 (1 - \frac{20}{2 \times 5}) = - 200 cm = - 2 m$

$(b)$ If the distance of the particle from the origin is $10 cm$ , then its $x$ coordinate is either $10 cm$ or $- 10 cm$ .

Hence, substituting $x = 10 cm$ in the equation $(2)$ , we get,

$10 = 10 t (1 - \frac{t}{10})$

$\Rightarrow t^{2} - 10 t + 10 = 0$
$\Rightarrow t = \frac{10 \pm \sqrt{100 - 40}}{2} \Rightarrow t = 5 \pm \sqrt{15} s$

Now, substituting $x = - 10 cm$ in equation $(2)$ , we get,
$- 10 = 10 t (1 - \frac{t}{10})$

$\Rightarrow t^{2} - 10 t - 10 = 0$

$\Rightarrow t = \frac{10 \pm \sqrt{100 + 40}}{2} \Rightarrow t = (5 \pm \sqrt{35}) s$

On solving, $t = 5 \pm \sqrt{35} s$ .
Neglecting negative value of time, we get,

$t = (5 + \sqrt{35}) s$

Hence, required moments of time are
$t = (5 \pm \sqrt{15}) s & (5 + \sqrt{35}) s$ .

$(c)$ Let us calculate the time when particle takes turn, i.e., the particle momentarily comes to rest. For rest,

$v = 0$

$\Rightarrow v_{0} (1 - \frac{t}{τ}) = 0 \Rightarrow t = τ \Rightarrow t = 5.0 s$

$x -$ coordinate of the particle at $t = 5.0 s$ is,

$x = 10 \times 5 (1 - \frac{5}{10}) = 25 cm$ .

Hence, the particle moves from $x = 0$ to $x = 25 cm$ in the first $5.0 s$ , then it take turns and start moving along negative $x$ -axis.

Now, $x$ -coordinate of the particle at $t = 4.0 s$ is,

$x = 10 \times 4 (1 - \frac{4}{10}) = 24 cm$ .

And, $x$ -coordinate of the particle at $t = 8.0 s$ is,

$x = 10 \times 8 (1 - \frac{8}{10}) = 16 cm$ .

Keeping these points in mind we can plot position vs. time graph as

Hence, the distance travelled by the particle in first $4.0 s$ is $24 cm$ , and the distance travelled by the particle in $8.0 s$ is $[25 + (25 - 16)] cm = 34 cm$ .

As the distance never decreases, so we can plot position vs. time graph as

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