Question

A particle moves in a straight line with the velocity as shown in the figure. At $t=0,x=-16\text{m,}$ then,

• A. The maximum value of the position coordinate of the particle is $54\text{m}$
• B. The maximum value of the position coordination of the particle is $36\text{m}$
• C. The particle is at the position of $36\text{m}$ at $t=18\text{s}$
• D. The particle is at the position of $36\text{m}$ at $t=30\text{s}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The maximum value of the position coordinate of the particle is $54\text{m}$
C The particle is at the position of $36\text{m}$ at $t=18\text{s}$
D The particle is at the position of $36\text{m}$ at $t=30\text{s}$

Maximum value of position coordinate = initial coordinate + area under the graph upto $t=24\text{s}$ (As upto $t=24\text{s}$, the displacement of the particle will be positive )
$x_{max}=-16+\left[\right(2\times 10)+\left(\frac{2+6}{2}\right)\times (18-10)+\frac{1\times 6}{2}\times (24-18\left)\right]$
$=-16+[20+32+18]=54\text{m}$

At time $t=18\text{s}$
$x_{t=18}=-16+$Area of graph upto $t=18\text{s}$
$x_{t=18}=-16+[20+32]=36\text{m}$

At time $t=30\text{s}$
$x_{t=30}=-16+$Area of graph upto $t=30\text{s}$
$x_{t=30}=-16+[70-\frac{1}{2}\times 6\times 6]=36\text{m}$