Answer:
Acceleration of the object is [tex]4\ m/s^2[/tex].
Explanation:
It is given that, the position of the object is given by :
[tex]r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j[/tex]
Velocity of the object, [tex]v=\dfrac{dr}{dt}[/tex]
Acceleration of the object is given by :
[tex]a=\dfrac{d^2r}{dt^2}[/tex]
[tex]a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)[/tex]
Using the property of differentiation, we get :
[tex]a=\dfrac{d^2r}{dt^2}=-4\ m/s^2[/tex]
So, the magnitude of the acceleration of the object at time t = 2.00 s is [tex]4\ m/s^2[/tex]. Hence, this is the required solution.
The magnitude of the acceleration of the object at time t = 2.00 s is [tex]-4 \;\rm m/s^{2}[/tex].
Given data:
The position of object is, [tex]r= [2.0 \;\rm m+(5.00 \;\rm m/s)t]i+[3.0 \;\rm m-(2.00 \;\rm m/s^{2})]j[/tex].
The time interval is, t = 2.00 s.
The velocity of object is obtained by differentiating the position with respect to time and acceleration is obtained by differentiating the velocity with time.
Velocity of the object is,
v = dr/dt
And acceleration of the object is given as,
[tex]a=\dfrac{d^{2}r}{dt}\\\\a=\dfrac{d^{2}[2.0 \;\rm m+(5.00 \;\rm m/s)t]i+[3.0 \;\rm m-(2.00 \;\rm m/s^{2})]j}{dt}\\\\a = -4 \;\rm m/s^{2}[/tex]
Thus, we can conclude that the magnitude of the acceleration of the object at time t = 2.00 s is [tex]-4 \;\rm m/s^{2}[/tex].
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