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Grade: 12th pass
        
A particle of mass 3kg is moving along x-axis and its position at time t is given by eqn x=(2t
11 months ago

Answers : (1)

Devendra Dansena
19 Points
							By the given eqn x=2t^2+5 We can get the acceleration by double differentiating the eqn d^2x/dt^2 = acceleration= 4m/s^2  Now by 2nd low of equation S=ut+1/2at^2 Let initial velocity be = 0 Then we getS=1/2*4*9= 18m         ( t=3)Now work done = FS =m*a*sW= 3*4*18= 216 joules
						
11 months ago
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