Question

An object has a position given by the radius vector r = [2.0 m + (3.00 m/s)t](i)+ [3.0 m - (2.00 m/s^2)t^2](j). Here (i) and (j) are the unit vectors along x and y and all quantities are in SI units. What is the speed and magnitude of the acceleration in m/s^2 of the object at time t = 2.00 s

Answer 1

Answer:

The speed of the object is ($3i - 4.00tj$)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 ($v$), take the first derivative of equation (i) with respect to time t as follows;

$v$ = $\frac{dr}{dt}$

 $v$ = $\frac{d[(2.0 + 3.00t)i + (3.0 - 2.00t^2)j] }{dt}$  

$v$  = $3i - 4.00tj$      ------------------------(ii)

To get the acceleration vector ($a$), take the first derivative of the speed vector in equation(ii) as follows;

$a = \frac{dv}{dt}$

$a = \frac{d(3i - 4.00tj)}{dt}$

$a = -4.00$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 ($3i - 4.00tj$)m/s

the magnitude of the acceleration is 4.00m/s²