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# The Position of a Particle is Given by 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? and What is the Magnitude and Direction of Velocity of the Particle at T = 2.0 S - Physics

The position of a particle is given by

r = 3.0t  hati − 2.0t 2  hatj + 4.0  hatk m

Where is in seconds and the coefficients have the proper units for to be in metres.

(a) Find the and of the particle?

(b) What is the magnitude and direction of velocity of the particle at = 2.0 s?

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#### 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`

Is there an error in this question or solution?
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#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 4 Motion in a Plane
Q 20 | Page 87
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#### Video TutorialsVIEW ALL [1]

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