Question

A velocity - time graph is shown above in figure (i) and (ii) find the acceleration and displacement.

Answer 1

Figure (i) represents a velocity-time graph when the body starts from rest, and its velocity increases at a uniform rate.
The slope of the graph AC. i.e., $\frac{AB}{BC}$ gives the acceleration of body.
$\therefore Acceleration=\frac{changeinvelocity}{Totaltime}$
$=\frac{(16-0)m{s}^{-1}}{5s}=3.2m{s}^{-2}$
The area of triangle ABC, gives the displacement.
Thus, displacement in 5 s
$\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 16m{s}^{-1}\times 5s$
$=40m$
Figure (ii) represents velocity - time graph, where the body is initially not at rest.
The slope of graph AC, i.e., $\left(\frac{AB}{BC}\right)$ gives the acceleration.
$\therefore Acceleration=\frac{AB}{BC}\frac{(25-5)m{s}^{-1}}{4s}=\frac{20}{4}m{s}^{-2}$
$=5m{s}^{-2}$
The distance covered by the body in specified direction is area of trapezium ECAD.
$\therefore Displacement=\frac{1}{2}(CE+AD)\times ED$
$=\frac{1}{2}(5+25)m{s}^{-1}\times 4s=60m$