Question

An object has a position given by r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s2)t2] j^, where all quantities are in SI units. What is the magnitude of the acceleration of the object at time t = 2.00 s?

Answer 1

Answer:

Acceleration of the object is $4\ m/s^2$.

Explanation:

It is given that, the position of the object is given by :

$r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j$

Velocity of the object, $v=\dfrac{dr}{dt}$

Acceleration of the object is given by :

$a=\dfrac{d^2r}{dt^2}$

$a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)$

Using the property of differentiation, we get :

$a=\dfrac{d^2r}{dt^2}=-4\ m/s^2$

So, the magnitude of the acceleration of the object at time t = 2.00 s is $4\ m/s^2$. Hence, this is the required solution.