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

The magnitude of the acceleration of the object at time t = 2.00 s is 6.40 m/s^2.

Step-by-step explanation:

To find the magnitude of the acceleration of the object at time t = 2.00 s, we can find the derivative of the position function with respect to time twice. The given position function is r = (2.0 m + (5.00 m/s)t)i + (3.0m−(2.00 m/s^2)t^2) j. Taking the derivative twice, we get a(t) = 5.00i + (-4.00 j).

Therefore, the magnitude of the acceleration of the object at time t = 2.00 s is |a(2.00 s)| = √(5.00^2+(-4.00)^2) = 6.40 m/s^2.