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lesya692 [45]
3 years ago
6

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?
Physics
1 answer:
makkiz [27]3 years ago
5 0

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.

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