Question

If velocity depends on time such that $v=4-2t$. Find out distance travelled by particle from $1$ to $3\text{s}$.

• A. Zero
• B. $2\text{m}$
• C. $3\text{m}$
• D. $4\text{m}$

Answer 1

The correct option is B $2\text{m}$

In order to calculate net distance we need to integrate for total path length travelled during the given time interval.

From the equation, $v=4-2t$,

It is clear that at $t=2\text{s}$ particle stops and after that particle's velocity becomes negative, that means the particle takes U- turn.

So, the net distance $d$ in between $t=1\text{s}$ to $t=3\text{s}$ is given by

$d=\left|{\int }_{t=1}^{2}vdt\right|+\left|{\int }_{t=2}^{t=3}vdt\right|$

$\Rightarrow d=\left|{\int }_1^2(4-2t)dt\right|+\left|{\int }_2^3(2t-4)dt\right|$

$\Rightarrow d={\left|(4t-t^2)\right|}_1^2+{\left|(t^2-4t)\right|}_2^3$

$\Rightarrow d=\left|4\right(2-1)-(2^2-1^2\left)\right|+\left|\right(3^2-2^2)-4(3-2\left)\right|$

$\Rightarrow d=1+1=2\text{m}$

Hence, option $\left(b\right)$ is correct.