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`        A body of mass 10 kg is being acted upon a force 3t^2 and an opposing constant force of 32N. the initial speed is 10m/s. Find velocity of the body after 5 sec.`
6 years ago

510 Points
```										net force on particle  = (3t2-32)N
net force = ma
ma = 3t2-32
10a = 3t2-32
a = 1/10 (3t2-32)
now a = dv/dt
dv/dt = 1/10 (3t2-32)
dv = 1/10 (3t2-32)dt
vf - u = 1/10 (3(t3/3) - 32t)            lim 0 to 5
vf = u + 1/10 (t3-32t)
at t=5 & u =10m/s
vf = 10 - 3.5  = 6.5m/s
so velocity  of particle is 6.5m/s .....
approve my ans if u like it
```
6 years ago
Technical Hacks
29 Points
```										 net force on particle  = (3t2-32)N net force = ma ma = 3t2-32 10a = 3t2-32 a = 1/10 (3t2-32)       now a = dv/dt   dv/dt = 1/10 (3t2-32)  dv = 1/10 (3t2-32)dt vf - u = 1/10 (3(t3/3) - 32t)            lim 0 to 5 vf = u + 1/10 (t3-32t) at t=5 & u =10m/s vf = 10 - 3.5 = 6.5m/s so velocity  of particle is 6.5m/s ..... approve my ans if u like itthanks highfive
```
one month ago
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