Question

The position of a particle is given by $r = 3.0t\widehat i + 2.0{t^2}\widehat j + 5.0\widehat k$ , where ‘$t$’ is in second and the coefficients have the proper units for ‘$r$’to be in metre. Find $u(t)$ and $a(t)$ of the particle.

Answer 1

Hint We know that velocity is equal to rate of change of position with respect to time.
i.e., $\overrightarrow v = \dfrac{{d\overrightarrow r }}{{dt}}$
where, $\overrightarrow r $ is position vector.
Now, we know acceleration is the rate of change of velocity with respect to time.
i.e., $\overrightarrow a = \dfrac{{d\overrightarrow v }}{{dt}}$
where, $\overrightarrow v $ is velocity vector.

Complete Step by step solution
Given: the position vector = $r = 3.0t\widehat i + 2.0{t^2}\widehat j + 5.0\widehat k$
We know velocity is the rate of change of position with respect to time.
Therefore, we have
$
  u(t) = \dfrac{{dr}}{{dt}} \\
  u(t) = \dfrac{{d\left( {3.0t\widehat i + 2.0{t^2}\widehat j + 5.0\widehat k} \right)}}{{dt}} \\
  u(t) = 3.0\widehat i + 4t\widehat j + 0 \\
  u(t) = 3.0\widehat i + 4t\widehat j \\
 $
Hence, the required velocity of the particle is $u(t) = 3.0\widehat i + 4t\widehat j$
Now we know that the acceleration is rate of change of velocity.
Therefore, we have
$a(t) = \dfrac{{du(t)}}{{dt}}$
Using the above value of $u(t)$ we get
$
  a(t) = \dfrac{{d(3.0\widehat i + 4t\widehat j)}}{{dt}} \\
  a(t) = 0 + 4\widehat j \\
  a(t) = 4\widehat j \\
 $

Hence, the required value of acceleration is $a(t) = 4\widehat j$

Note The case of constant acceleration and the motion in a straight line yields some simple equation that permits the evaluation of the velocity and the position of the vehicle if the initial conditions are known. From the definition, we know $a = \dfrac{{dv}}{{dt}}$ , the velocity at later time $t$ can be determined from the initial velocity, $v(0)$ , and the constant acceleration $a$ , by integration. This gives
$v(t) = v(0) + at$
Similarly, on implementing different conditions of position and velocity we get
$
  s(t) = v(0) + \dfrac{1}{2}a{t^2} \\
  {v^2}(t) = {v^2}(0) + 2as \\
 $