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Grade: 12th pass

                        

The velocity v of a particle moving along x axis varies with its position x as shown in figure. The acceleration of a particle varies with position as

3 years ago

Answers : (3)

Eshan
askIITians Faculty
2095 Points
							Dear student,

Acceleration of a particle=a=\dfrac{dv}{dt}=v\dfrac{dv}{dx}

Velocity as a function of displacement can be writtten from the given figure as-

v=4-2x
\implies \dfrac{dv}{dx}=-2

Hencea=v\dfrac{dv}{dx}=(4-2x)(-2)=4x-8

2 years ago
Rajat
13 Points
							
a==dv/dt
dv/dx×dx/dt
a=vdv/dx
From graph 
dv/dx=2
a=-2v........1
- indicates line of graph
Now using eqn of slope
y=mx+c
v=-2+4
- is due to negative slope 
a=-2(-2x+4)
a=4x-8
 
one year ago
Kushagra Madhukar
askIITians Faculty
605 Points
							
Dear student,
Please find the solution to your problem.
 
a = dv/dt
dv/dx × dx/dt
a = vdv/dx
From graph 
dv/dx = 2
a = – 2v   ........1
 – sign indicates line of graph
Now using eqn of slope
y = mx + c
v = – 2x + 4
 – sign is due to negative slope 
a = – 2(–2x + 4)
a = 4x – 8
 
Thanks and regards,
Kushagra
one month ago
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