Question

A particle is moving along x-y plane. Its x and y co-ordinates very with time as x=2t^2 and y=t^3 Here, x abd y are in metres and t in seconds. Find average acceleration between a time interval from t=0 to t=2 s.

Answer 1

The average acceleration is $a _{avg} =(4 \vec i + 6 \vec j )$ $\frac{m}{s^{2} }$.

Explanation:

Given :

The co-ordinate of X-axis = $2t^{2} \vec i$

The co-ordinate of Y-axis = $t^{3} \vec j$

The position vector is given by,

     $r = 2t^{2} \vec i + t^{3} \vec j$

From the formula of velocity,

    $v = \frac{\vec dr}{dt}$

    $v = 4t \vec i + 3t^{2} \vec j$

And average acceleration is given by,

    $a _{avg} = \frac{\Delta v}{\Delta t}$

    $a _{avg} = \frac{v_{2 sec} - v_{0 sec} }{2 - 0}$

   $a _{avg} =\frac{ [4(2) \vec i +3 (2)^{2} \vec j ]}{2}$

   $a _{avg} =(4 \vec i + 6 \vec j )$ $\frac{m}{s^{2} }$