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A particle A of mass 10/7 kg is moving in the positive direction of x.Its initial position is x=0 & initial velocity is 1m/s.The velocity at x=10 is?

A particle A of mass 10/7 kg is moving in the positive direction of x.Its initial position is x=0 & initial velocity is 1m/s.The velocity at x=10 is?

Grade:12

9 Answers

Suraj Tripathi
18 Points
9 years ago

the velocity at x = 0 will be 1m/s as you have not mentioned any acceleration...

dinesh sharma
37 Points
9 years ago

without accn. this impossible to predict about velocity.

Gaurav
13 Points
4 years ago
The answer is possible if the velocity will differentiate w.r.t time and hence we got tthe acceleration by using eqn. S=ut+1/2at×t
dinesh sharma
37 Points
4 years ago
Well...we dont know the acceleration is constant or variable and this is the real problem.so its impossible to predict the motion.
Akhil Jaywant
24 Points
4 years ago
Chutiye graph to de hutiye..... Graph bhii hotaa h is ques me......................................................
Basma
23 Points
4 years ago
This question cannot be solved as it is not given whether it is uniform motion or non uniform motion or accelarated motion.the mass is not needed to find velocity.
Amarjit
105 Points
4 years ago
If there is no acceleration in x direction then its velocity will remain throughout the motion, hence velocityvat x=10 will be 1m/s.
Mohit
13 Points
4 years ago
This graph contains a graph which is a straight line ... Graph is in between power ( watt) and disp (m)The graphs starts from (0,2) and end on (10,4)By this find relation between power and displacement and integrate it by using formulae p = force × velocity And put force = mass and acc.And acc = v dv/dx and integrate it Thanks
Shubham Patil
15 Points
3 years ago
Area under P-x graph
∫pdx=∫(m  d/dt)vdx
∫mv^2dv=[mv^3/3]:(limits 1 to v both sides)
=10/7×1/3(v^3-1)
From graph:area=1/2(2+4)×10=30
Therefore 10/3×1/3(v^3-1)=30
v=4m/s
Hope it helps!!!
Cheers!!!
By Shubham Patil
 

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