Answer:
The speed of the object is ([tex]3i - 4.00tj[/tex])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 ([tex]v[/tex]), take the first derivative of equation (i) with respect to time t as follows;
[tex]v[/tex] = [tex]\frac{dr}{dt}[/tex]
[tex]v[/tex] = [tex]\frac{d[(2.0 + 3.00t)i + (3.0 - 2.00t^2)j] }{dt}[/tex]
[tex]v[/tex] = [tex]3i - 4.00tj[/tex] ------------------------(ii)
To get the acceleration vector ([tex]a[/tex]), take the first derivative of the speed vector in equation(ii) as follows;
[tex]a = \frac{dv}{dt}[/tex]
[tex]a = \frac{d(3i - 4.00tj)}{dt}[/tex]
[tex]a = -4.00[/tex]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 ([tex]3i - 4.00tj[/tex])m/s
the magnitude of the acceleration is 4.00m/s²
