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A particle moving in the xy-plane has velocity v⃗ =(2ti^+(3−t2)j^)m/s, where t is in s.

Part A

What is the x component of the particle's acceleration vector at t = 6 s?

Express your answer with the appropriate units.

Part B

What is the y component of the particle's acceleration vector at t = 6 s?

Express your answer with the appropriate units.

Answer :

Answer:

A)   2 m/s²

B) -12 m/s²

Explanation:

A)

  • Applying the definition of instantaneous acceleration as the derivative of the velocity with respect to time, the acceleration vector can be expressed as follows:
  • a = 2i^ + (-2t) j^ m/s²
  • So, the x-component of the acceleration vector is constant, and equal to 2m/s², at any time.
  • When t=6 s, aₓ= 2 m/s²

B)

  • Applying the definition of instantaneous acceleration as the derivative of the velocity with respect to time, the acceleration vector can be expressed as follows:
  • a = 2i^ + (-2t) j^ m/s²
  • So, when t= 6s, the y component of a is as follows:
  • ay = -2(6s) = -12 m/s²

A) The x-component of the particle's acceleration is; 2 m/s²

B) The y-component of the particle's acceleration at t = 6 is; -12 m/s²

X and Y components of Acceleration

We are given the velocity as;

v^ = 2t i^ + (3 − t²) j^ m/s

A) We know that acceleration is simply a derivative of velocity with respect to time. Thus;

a^ = dv/dt = 2 i^ + (-2t) j^ m/s²

The x - component of the acceleration is the figure attached to the vector i^.

Thus, x-component of acceleration = 2 m/s²

B) The y-component of the acceleration will be the value attached to the vector j^. Thus;

y-component of acceleration = -2t m/s²

At t = 6 s, we have;

y-component of acceleration = -2(6) = -12 m/s²

Read more about x and y components of acceleration at; https://brainly.com/question/7114558