Question

A particle is moving along the positive $x$-axis and at $t=0$, the particle is at $x=0$. The acceleration of the particle is a function of time. The acceleration at any time $t$ is given by $a=2(1–[t\left]\right)$ where $\left[t\right]$ is the greatest integer function. Assuming that the particle is at rest initially, the displacement of the particle in $4s$ is

• A. $1m$
• B. $2m$
• C. $6m$
• D. $4m$

Answer 1

The correct option is B $2m$

Given that at $t=0,x=0$ and accelaration $a=2(1-[t\left]\right)$ where $\left[t\right]$ is the greatest integer function.
$\frac{dv}{dt}=2-2\left[t\right]$
$\Rightarrow {\int }_0^{v}dv={\int }_0^4(2-2[t\left]\right)dt$
$\Rightarrow v={\int }_0^4(2-2[t\left]\right)dt$
$\text{For}t=0\text{to}t=1s,\left[t\right]=0,v={\int }_0^{1}2dt=2m/s=v_1$
$\text{For}t=1s\text{to}t=2s,\left[t\right]=1,v_2-v_1={\int }_1^{2}0dt=0\Rightarrow v_1=v_2=2m/s$
$\text{For}t=2s\text{to}t=3s,\left[t\right]=2,v_3-v_2={\int }_2^3-2dt=-2m/s\Rightarrow v_3=-2+2=0m/s$
$\text{For}t=3\text{to}t=4s,\left[t\right]=3,v_4-v_3={\int }_3^4-4dt=-4m/s=v_4$



Displacement is given by area under the $v-t$ graph
= Area above $x$-axis - Area below $x$-axis
$=\frac{1}{2}(3+1)\times 2-\frac{1}{2}\times 4\times 1=2m$
$\therefore \text{Displacement in}4s=2m$