Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
171 views
in Physics by (85.7k points)
closed
The velocity - displacement graph of a particle moving along a straight line is shown
The most suitable acceleration - displacement graph will be
image
A. image
B. image
C. image
D. image

1 Answer

0 votes
by (90.5k points)
 
Best answer
Correct Answer - A
The equation for the given ` v-x` graph is
` v = - (v_(0))/(x_(0)) + v_(0)` ….. (i)
`(dv)/(dx) = - ( v^(0))/(x_(0))`
`:. v(dv)/(dx) = - (v)/(x_(0)) xx v = -(v_(0))/(x_(0))[-(v_(0))/(x_(0))+ v_(0)] ` from (1)
`:. a = (v_(0)^(2))/(x_(0)^(2))x - (v_(0)^(2))/(x_(0))` ....(ii) `[ because a = v(dv)/(dx)]`
On comparing the equation (ii) with equation of a straight line
` y = mx + c`
We get`m = (v_(0)^(2))/(x_(0)^(2)) = +ve ` ,
i.e. ` tan theta = + ve` , i.e., `theta ` is acute.
Also `c = -(v_(0)^(2))/(x_(0)^(2))` ,
i.e., the `y - intercept ` is negative
The above conditions are satisfied in graph `(a)`.
image

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...