Question

A particle moves along $x$ -axis with an initial speed $v_{0} = 5 {m s}^{- 1}$ . If its acceleration varies with time as shown in $a - t$ graph in the figure,

(a) Find| the velocity of the particle at $t = 4 s$ .

(b) Find the time when the particle starts moving along $- x$ -direction.

Accepted Answer

The velocity of the particle at $t = 4 s$ can be given as,

${\vec{v}}_{4} = {\vec{v}}_{0} + Δ \vec{v} \dots (1)$

where $Δ \vec{v} \equiv A$ ( $=$ area under $a - t$ graph during first $4 s$ )

Referring to $a - t$ graph (figure), we have,
$A = A_{1} + A_{2} - A_{3} - A_{4} \dots (2)$

where $A_{1} = 5 \times 1 = 5, A_{2} = \frac{1}{2} \times x \times 5$ , $A_{3} = \frac{1}{2} \times (1 - x) \times 10$ and $A_{4} = \frac{1}{2} \times 2 \times 10 = 10$

We can find $x$ as follows:

Using properties of similar triangles, we have $\frac{x}{5} = \frac{1 - x}{10}$ . This yields $x = \frac{1}{3}$ .

Substituting $x = \frac{1}{3}$ in $A_{2} = \frac{1}{2} \times x \times 5$ and $A_{3} = \frac{1}{2} (1 - x) \times 10$ , we have $A_{2} = \frac{5}{6}$ and $A_{3} = \frac{10}{3}$ .

Then substituting $A_{1}, A_{2}, A_{3}$ and $A_{4}$ in $(2)$ , we have $A = - 7.5$ . The negative area tells us that change in velocity is along $- x$ -direction.

$Δ \vec{v} = - 7.5 m s^{- 1}$

Hence, substituting in $(1)$ ,

${\vec{v}}_{0} = 5 {m s}^{- 1}$ and $Δ \vec{v} = - 7.5 {m s}^{- 1}$

we have,

${\vec{v}}_{4} = {\vec{v}}_{0} + Δ \vec{v} = 5 - 7.5 = - 2.5 m s^{- 1} .$

B M Sharma Solutions for Chapter: Kinematics I, Exercise 63: ILLUSTRATION 4.59

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 63: ILLUSTRATION 4.59

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 63: ILLUSTRATION 4.59 with Hints & Solutions

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