Answer:
The speed of the object is ()m/s
The magnitude of the acceleration is 4.00m/s²
Explanation:
Given - position vector;
r = (2.0 + 3.00t)i + (3.0 - 2.00t²)j -------------------(i)
To get the speed vector (), take the first derivative of equation (i) with respect to time t as follows;
=
=
=
------------------------(ii)
To get the acceleration vector (), take the first derivative of the speed vector in equation(ii) as follows;
j
The magnitude of the acceleration |a| is therefore given by
|a| = |-4.00|
|a| = 4.00 m/s²
In conclusion;
the speed of the object is ()m/s
the magnitude of the acceleration is 4.00m/s²