In the figure the variations of componets of acceleration of particles of mass 1kg is shown w.r.t. time. The initial velocity of the particle is `vec(u)=(-3 hati+4hatj)`m/s. the total work done by the resultant force on the particles in time intervals from t=0 to t=4 seconds is :

Options

A. `22.5 J`
B. 10 J
C. 0
D. None of these

Answer

B

Explanation

From given graphs :
`a_(x) =(3)/(4)t` and `a_(y) =-((3)/(4)t+1) rArr v_(x) =(3)/(8) t^(2) +C`
At `t =0 ,v_(x) = -3 rArr C = -3`
:. `v_(x)=(3)/(8)t^(2) -3 rArrdx =((3)/(8)t^(2)-3) dt` ….(1)
Similarly , `dy =(-(3)/(8)t^(2)-t+4)dt` ….(2)
As `dw =vec F. vecds = vec F.(dx i+xy j)`
`:. underset(0) overset(w) int dw = underset(0) overset(4) int[(3)/(4) ti-((3)/(4) t+1)j].x`
`[((3)/(8) t^(2)-3)i+(-(3)/(8)t^(2)-t+4) j]dt`
`:. W = 10 J`
Alternate Solution :
Area of the graph ,
`int a_(x) dt =6=V_((x)f)-(-3) rArr V_((x)f) = 3`
and `int a_(y) dt =-10=V_((y)f)-(-4)rArr V_((y)f) = -6`
Now work done `= Delta KE = 10 J`.