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A particle A is moving in xy plane with the constant speed of 2pi m//s along the path x^(2)+y^(2)+120y=0. At time t=0. When A is at the origin, another particle B starts moving from origin with constant acceleration is such a way that at time t=5s velocities of both the particles are found to be equal. If sense of motion of A is clockwise, calculate the average speed of B over first five second. |
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Answer» `2pi m//s` `theta=pi//6` `vecV_(A)=-v_(0)cos30^(@)hati-v_(0)SIN30^(@)hatj` `=-(v_(0))/(2)[sqrt(3)hati+hatj]` `rArrvecV_(B)=vecV_(A)=-(V_(0))/(2)[sqrt(3)hati+hatj]` `veca_(B)=(-V_(0))/(2)[sqrt(3)hati+hatj]` Since particle starts moving with uniform acceleration so it will move on a straight line `vecS_(B)=vecU_(B)t+(1)/(2)veca_(B)t^(2)=0+(V_(0))/(20)[sqrt(3)hati+hatj]xx25` `|vecS_(B)|=` Distance travelled by particle `B=(2pi)/(20)xx25xx2=5pi` Average SPEED of particle `B` over first five second `(5pi)/(5)=pim//s` Second method `:' veca_(B)=`CONSTANT, to `0 lt V_(B) lt 5sec = |(vecV_(B(a))+vecV_(B(5)))/(2)|=(V_(0))/(2)=pim//s` 
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