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Grade: 12th pass
        
Please solve the question in the attachment...................... 
3 months ago

Answers : (3)

SARTHAK TOMAR
14 Points
							
The answer will be d bcoz first the acceleration is negative hence velocity decrease but does not become zero and after that the vice versa. Hope u get the answr
3 months ago
SARTHAK TOMAR
14 Points
							
The answer will be A because first the displacement is decreasing and therefore velocity should be negative and should increase upto t=1 but suddenly after t=1 the displacement starts to increase. Therefore we can say graph mentioned in (a) will be correct
3 months ago
Pratik Kumawat
40 Points
							
Hey Pawan joshi,
lets move towards the solution
In the first curve displacement(s) is decreasing with increase in time .
and you can see the curve is like a parabola in only first quadrant having eqn t2= -4as (refer notes of parabola).
differentiating it , we will get  ds/dt = -(1/4a)*2t , where ds/dt is velocity.
and it is negative and linearly dependent on time .
for curve between t=1 and t=2 ,
the slope of curve is decreasing linearly and it is always positive .
so Option A is correct.
 
 
 
 
Pratik Kumawat
IIT(ISM) Dhanbad
    
3 months ago
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  • Discussion Forum
  • Previous Year Exam Questions


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