Consider the position of a particle moving in three dimensional space to be given by the position vector r = 2.5 t 3 i - 10 t 2 j + 4.00 to, where r is meters and t in seconds. What is the magnitude of the acceleration of the particle at t = 1.0 s?

25m/s²

Explanation:

Given the position of a particle by the three dimensional vectors;

r = 2.5 t³ i - 10 t² j + 4.00t k where r is in meters and r in seconds, we will get the magnitude of the velocity(v) first since velocity is rate of change in displacement.

v = dr/dt where r is the position or displacement of an object

Velocity is gotten by differentiating the given function with respect to time to have;

v = 7.5t² i - 20t j + 4 k

Acceleration (a) is the rate of change in velocity.

a = dv/dt

Acceleration is derived by differentiating the velocity function with respect to time to have;

a = 15t i - 20j

Acceleration of the particle at t = 1.0s will be;

a = 15(1) i - 20j

a = 15i - 20j

Magnitude of the acceleration can be gotten by finding the resultant of the resulting vector

a = √(15i)²+(20j)²

Since i.i = j.j = 1

a = √15²+20²

a = √225+400

a = √625

a = 25m/s²

Therefore the magnitude of the acceleration of the particle at t = 1.0 s is 25m/s²

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