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प्रश्न
The position of a particle is given by
`r = 3.0t hati − 2.0t 2 hatj + 4.0 hatk m`
Where t is in seconds and the coefficients have the proper units for r to be in metres.
- Find the v and a of the particle?
- What is the magnitude and direction of velocity of the particle at t = 2.0 s?
उत्तर
Here `vecr(t) = (3.0t hati - 2.0t^2 hatj + 4.0 hatk)m`
(a) `vecv(t) = vec(dr)/(dt) = (3.0 hati - 4.0t hatj) "m/s"`
and `veca(t) = vec(dt)/(dt) = (-4.0 hatj)"m/s"^2`
(b) Magnitude of velocity at t= 2.0 s
`v_(t = 2s) = sqrt((3.0)^2+(-4.0xx2)^2) = sqrt(9+64) = sqrt(73)`
= 8.54 `ms^(-1)`
This velocity will subtend an angle `theta` from x-axis, where `tan theta =((-4.0xx2))/(3.0)` = -2.667
`:.theta = tan^(-1)(-2.6667) =-69.44^@ = 69.44^@` with x-axis
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