__Question 4.20__:

The position of a particle is given by

r = 3.0t

**i**− 2.0t

^{2}

**j**+ 4.0

**k**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__:

**v**(t) = (3.0

**i**- 4.0t

**j**)

**a**= -4.0

**j**

The position of the particle is given by:

**r**= 3.0t

**i**-2.0t

^{2}

**j**+4.0

**k**

**Velocity v of the particle is given as:**

**v**= d

**r**/ dt = d (3.0t

**i**-2.0t

^{2}

**j**+ 4.0

**k**) / dt

∴

**v**= 3.0

**i**-4.0t

**j**

**Acceleration**

**a**of the particle is given by:

**v**= (3

^{2}+ (-8)

^{2})

^{1/2}= (73)

^{1/2}= 8.54 m/s

Direction, θ = tan

^{-1}(v

_{y}/v

_{x})

= tan

^{-1}(-8/3) = -tan

^{-1}(2.667)

= -69.45

^{0}

The negative sign indicates that the direction of velocity is below the

*x*-axis.

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