The x and y positions (in meters)of a particle are given by: x(t)= 253 t,and y(t)= -12.0 t² + 155 t + 100.The magnitude of the particle's acceleration at t = 12.9 s is A)0 m/s². B)-12.0 m/s². C)12.0 m/s². D)-24.0 m/s². E)24.0 m/s².

Question

Quizplus · Accepted Answer

To find the acceleration of the particle, we need to take the second derivative of the position function with respect to time:

x(t) = 253t   (1)
y(t) = -12.0t^2 + 155t + 100   (2)

dx/dt = 253   (3)
dy/dt = -24t + 155   (4)

Taking the derivative of (3) and (4), we get:

d^2x/dt^2 = 0   (5)
d^2y/dt^2 = -24   (6)

Substituting t = 12.9s into (6), we get:

d^2y/dt^2 = -24 m/s^2

The magnitude of the acceleration is equal to the absolute value of the acceleration vector, which is equal to |-24 m/s^2| or 24 m/s^2.

Therefore, the answer is E) 24.0 m/s^2.

The X and Y Positions (In Meters)of a Particle Are