Question

The position of a particle is given by r = 3.0 t i - 2.0t² 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 the velocity of the particle at t = 2.0 s?

Answer 1

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 -2.0t2 j +4.0 k

Velocity v of the particle is given as:

v = dr / dt = d (3.0t i -2.0t2 j + 4.0 k) / dt

∴ v = 3.0 i -4.0t j

Acceleration a of the particle is given by:

v = (32 + (-8)2)1/2 = (73)1/2 = 8.54 m/s

Direction, θ = tan-1(vy/vx)

= tan-1(-8/3) = -tan-1(2.667)

= -69.450

The negative sign indicates that the direction of velocity is below the x-axis.

Answer 2

Answer:

a) The velocity of the particle v = $3.0 i - 4.0 tj$ m/s and the acceleration of the particle is a = $-4.0j$ m/s^2.
b)The velocity of the particle at t=$2.0 sec$ is v=$3.0i-8.0j$ m/s

Explanation:

• In the question we are given the equation of position of particle as x = $3.0 t i - 2.0t^2 j + 4.0 k$ m where t is in seconds.
• Now as we know relation between velocity and distance when the equation of distance is given in terms of t is
  $V = \frac{\partial x }{\partial t}$ and relation between acceleration and distance when the equation of distance is given in terms of t is  is
  $a = \frac{\partial^2 x }{\partial t^2}$
• By putting the value of x in the above equations we get the value of velocity as v = $3.0 i - 4.0 tj$ m/s and the value of acceleration a = $-4.0j$ m/sec^2
• To find out the velocity at t = $2.0$ seconds we put the value of t in the equation v = $3.0 i - 4.0 tj$ m/s and we get the value of v as $3.0i-8.0j$ m/s
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