Question

Velocity of a particle travelling along a straight line represented by $+ve$ $x$-axis is given by
$4$ $0\le t<5$
$v\text{(m/s)}$ $=$ $2t$ $5\le t<10$
$6t^2$ $10\le t\le 20$
Find out the distance travelled by the particle between $t=4\text{s}$ to $t=12\text{s}$.

• A. $1505\text{m}$
• B. $1800\text{m}$
• C. $2100\text{m}$
• D. $860\text{m}$

Answer 1

The correct option is A $1505\text{m}$

In case of $1\text{D}$ motion;
$\text{Displacement}=\text{Distance}$

$\Rightarrow$Particle is not reversing it's direction of motion, because by putting any value of time in differentiation of $v$ w.r.t $t$ we get acceleration $a=+ve$

$\Rightarrow$ For $t=4\text{s}\text{to}t=5\text{s}$:
$v=4\text{m/s}$

For $t=6\text{s}\text{to}t=9\text{s}$:
$v=2t\text{m/s}$

For $t=10\text{s}\text{to}t=12\text{s}$:
$v=6t^2\text{m/s}$

Applying the equation:

$v=\frac{ds}{dt}$

$ds=vdt$

$\int ds={\int }_4^{12}vdt$

$\int ds={\int }_4^{5}4.dt+{\int }_6^{9}2t.dt+{\int }_{10}^{12}6t^2.dt$

$s={\left[4t\right]}_4^5+{\left[\frac{2t^2}{2}\right]}_6^9+{\left[\frac{6t^3}{3}\right]}_{10}^{12}$

$\Rightarrow s=4+[81-36]+2[{12}^3-{10}^3]$

$\therefore s=1505\text{m}$