Question

Referring to the $v^{2} - s$ diagram of a particle, find the displacement of the particle during the last $2 s$ .

Accepted Answer

Slope of $v^{2} - s$ graph is given by,

$m = - \frac{(2000 - 1500)}{100 - 50} = - 10$

Relation between $v^{2}$ and $s$ will be,

$v^{2} = - 10 s + C$ ; where $C$ in intercept of the graph.

At $s = 50, v^{2} = 2000$
$\Rightarrow 2000 = - 10 \times 50 + C$

$\Rightarrow C = 2500$

So, the relation between $v^{2}$ and $s$ will be,
$v^{2} = - 10 s + 2500 \dots (1)$

Differentiating equation $(1)$ both sides with respect to time,

$2 v \frac{d v}{d t} = - 10 \frac{d s}{d t}$

Here, $\frac{d v}{d t}$ is acceleration and $\frac{d s}{d t} = v$ , therefore,
$\Rightarrow 2 v a = - 10 v$

$\Rightarrow a = - 5 m s^{- 2}$ , means acceleration is constant and negative, so the motion will stop after some time.

To find initial velocity:
Put at $s = 0$ and let $v = u$ , then
$\Rightarrow u^{2} = 2500$

$\Rightarrow u = 50 {m s}^{- 1}$ will be the initial velocity.

Let at time $t = T$ , the motion of the particle is stopped, so from equation of motion,

$v = u + a t$

$\Rightarrow 0 = 50 - 5 T$
$\Rightarrow T = 10 s$ time of motion
Displacement during the last $2 s$ will be,
$S = S_{t = 10 s} - S_{t = 8 s}$

$S = [50 \times 10 - \frac{1}{2} 5 (10)^{2}] - [50 \times 8 - \frac{1}{2} 5 (8)^{2}] = 10 m$

B M Sharma Solutions for Chapter: Kinematics I, Exercise 71: ILLUSTRATION 4.68

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 71: ILLUSTRATION 4.68

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 71: ILLUSTRATION 4.68 with Hints & Solutions

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