Answer:
25m/s²
Step-by-step 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