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.
(a) Find the v and a of the particle?
(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?
Solution 1
(a)vecv(t) = (3.0 hati - 4.0t hatj); veca = -4.0 hatj
The position of the particle is given by:
`vecr =3.0t hati - 2.0t^2 hatj + 4.0 hatk`
Velocity `vecv`, of the particle is given as:
`vecv = (dvecr)/(dt) = d/(dt)(3.0t hati - 2.0t^2 hatj + 4.0 hatk)`
`:.vecv = 3.0 hati - 4.0t hatj`
Acceleration `veca`, of the particle is given as:
`veca = (dvecv)/(dt) = d/(dt)(3.0 hati - 4.0t hatj)`
`:.veca = -4.0 hatj`
(b) 8.54 m/s, 69.45° below the x-axis
We have velocity `vecv = 3.0 hati - 4.0t hatj`
At t = 2.0 s:
`vecv = 3.0 hati - 8.0 hatj`
The magnitude of velocity is given by:
|vecv|=sqrt(362+(-8)^2) = sqrt73 = 8.54 m/s
Direction, `theta = tan^(-1)(v_y)/v_x)`
= `tan^(-1)(-8/3) = -tan^(-1)(2.667)`
`=-69.45^@`
The negative sign indicates that the direction of velocity is below the x-axis.
Solution 2
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 `beta` from x-axis, where `tan beta =((-4.0xx2))/(3.0)` = -2.667
`:.beta = tan^(-1)(-2.6667) =- -69.44^@ = 69.44^@` from negative x-axis`
