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 correct graph to show the variation of velocity versus time for the interval $0<t<4s$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Given that at $t=0,x=0$ and accelaration $a=2(1-[t\left]\right)$ where $\left[t\right]$ is the greatest integer function.
$\frac{dv}{dt}=2-2\left[t\right]$
$\Rightarrow {\int }_0^{v}dv={\int }_0^4(2-2[t\left]\right)dt$
$\Rightarrow v={\int }_0^4(2-2[t\left]\right)dt$
$\text{For}t=0\text{to}t=1s,\left[t\right]=0,v=\int 2dt=2t$
$\text{For}t=1s\text{to}t=2s,\left[t\right]=1,v=\int 0dt=\text{constant}$
$\text{For}t=2s\text{to}t=3s,\left[t\right]=2,v=\int -2dt=-2t$
$\text{For}t=3\text{to}t=4s,\left[t\right]=3,v=\int -4dt=-4t$
Hence velocity of particle from $0$ to $1s$ is increasing, from $1s$ to $2s$ velocity is constant, from $2s$ to $3s$ velocity is decreasing and from $3s$ to $4s$ velocity is also decreasing.
Hence correct graph will be (c)