#### Question

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`