The velocity v versus time t graph for the motion of a particle is shown in the figure. Which of the following statement(s) is (are) correct [One or more option(s) correct]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$υ_{a v g} < 0.50 m/s for 0 s \leq t \leq 3 s$

$υ_{a v g} = 0.75 m/s for 3 s \leq t \leq 5 s$

$υ_{a v g} = 0.60 m/s for 0 s \leq t \leq 5 s$

acceleration is minimum at $t = 4 s$

answer is A, B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The area under the velocity-time graph gives the dis-placement s. The area under the curve OAB is less then the area of the triangle OBF. The area of the triangle OBF is 1.5 m. Thus, motion of the particle in the time interval $0 \leq t \leq 3 s,$
   $υ_{a v g, 1} = \frac{s_{1}}{t_{1}} = \frac{Area under OAB}{3} < \frac{1.5}{3} = 0.5 m/s.$
Question Image
Now consider the motion in the time interval $3 s \leq t \leq 5 s$ . The area of the shaded portions BCE and CDG are equal. BCE is below the curve BC (positive area) and CDG is above the curve CD (negative area). Thus, area under the curve BCD is equal to the area of rectangle EGHF; which is $s_{2} = 1.5 m .$ Thus, average velocity in this time interval is,
   $υ_{a v g, 2} = \frac{s_{2}}{t_{2}} = \frac{Area under BCD}{2} = 0.75 m/s.$
You may note an important point here. The average velocity in this time interval is the average of initial velocity $υ_{i} = 1 m/s$ and final velocity $υ_{f} = 0.5 m/s$ i.e., $υ_{a v g} = (υ_{i} + υ_{f}) / 2.$ this is always true for uniform acceleration. This is also true for non-uniform acceleration if area between the curve and $(υ_{i} + υ_{f}) / 2$ on the left side is equal to area between these two on the right side.
For motion in the time interval $0 \leq t \leq 5 s,$
   $υ_{a v g} = \frac{s_{1} + s_{2}}{t_{1} + t_{2}} = \frac{υ_{a v g, 1} t_{1} + υ_{a v g, 2} t_{2}}{t_{1} + t_{2}}$
The particle accelerates (a > 0) for the first three seconds and decelerate (a < 0) for the next two seconds. The point C is inflection point. The slope has minimum value at this point. Hence, the acceleration is minimum at $t = 4 s .$
$< \frac{(0.5) (3) + (0.75) (2)}{3 + 2} = 0.6 m/s.$