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A force of 6N acts on a body of mass 1.5 Kg for 2 seconds. Assuming the body to be initially at rest, find the distance covered in 5sec after the force starts acting.

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2 years ago

```							acceleration of body for 2 seca = 6/1.5 = 4 m/s2velocity after 2 secv = a*t = 4*2 = 8 m/sDistance coververd in 5 secd = ( ½*4*22) + (8*3) = 8+24 = 32 mHope it clears.
```
2 years ago
```							F=ma 6=1.5 *aa=4m/s2distance covered when the body is at  rest = s= 1/2at2     (that is because u =0m/s)Therefore ,s=1/2 (4)(4)=8m
```
2 years ago
```							FORCE=mass*acc.---------1                                   (F=6N,,M=1.5kg,,t=5 sec)s=ut+1/2at^2                                               ( we are given that body is initially at rest so,ut=0)so,s=1/2 at^2----------------2by,1 equation                             6=1.5*aa=6/1.5=4m/s^2by using equation 21/2*4*5^22*25=50N
```
2 years ago
```							sorry
```
2 years ago
```							CASE-1Given.F=6Nm=1.5kgt=2 seconds*F=maF/m=a6/1.5=4mps,sq.Case-2T=5 secU=0mpsS=ut+1/2a(t,sq.)S=0×5+1/2×4×25S=50m.
```
2 years ago
```							F=6 NM=1.5 kgt=2sec ,,F=mu/t ,,u=ft/m=12/1.5 ,,S=Ur+1/2at^2S=12/1.5+ 1/2*6/1.5*(5)^2S=40+50S=90 m
```
2 years ago
```							We know ....F =ma 6=1.5 × a 6/1.5 =a a= 4 m/sec square We also know that .....v=u+at v = 0+4×2v =8 m / secWe are aware that S =ut+ 1/2 at square S = 0×5 +1/2 ×4×2 ×2 S=8m = 8m + the distance of the 3 sec= 8+ 3×8 =8+24=32m
```
4 months ago
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