Question

a particle is moving along the y-axis. the particle’s position as a function of time is given by y=αt3−βt ϕ, where α=1m/s3, β=4m/s, and ϕ=3m. what is the particle’s acceleration at time t=3.0s?

Answer 1

The particle's acceleration at time t = 3.0s is 18 m/s^2.

Step-by-step explanation:

To find the particle's acceleration at time t = 3.0s, we need to differentiate the position function twice with respect to time. Considering the position function is y = αt^3 - βtϕ, we differentiate it twice to find the acceleration:

a(t) = d^2y/dt^2

Let's differentiate the position function with respect to time:

dy/dt = 3αt^2 - βϕ

Now, differentiate again to find the acceleration:

a(t) = d^2y/dt^2 = d(3αt^2 - βϕ)/dt = 6αt

Substituting the values of α and t into the equation, we get the particle's acceleration at t = 3.0s:

a(3.0s) = 6α(3.0s) = 6(1m/s^3)(3.0s) = 18 m/s^2

Answer 2

The particle's acceleration at time t=3.0s is found by taking the second derivative of the position function with respect to time, resulting in an acceleration of 18 m/s².

Step-by-step explanation:

To determine the particle’s acceleration at time t = 3.0s for its given position function y = αt³ − βt φ, where α = 1 m/s³, β = 4 m/s, and φ = 3m, we need to compute the second derivative of the position function with respect to time. The first derivative of the position function with respect to time gives us the particle's velocity, and the second derivative gives us the acceleration.

The first derivative of the position function is:

Velocity v(t) = d(y)/dt = 3αt² - βφ

The second derivative of the position function is:

Acceleration a(t) = d(v)/dt = d²(y)/dt² = 6αt

Substituting α = 1 m/s³ and t = 3.0s into the acceleration function, we find:

Acceleration a(3.0s) = 6(1 m/s³)(3.0s) = 18 m/s²

Therefore, the particle's acceleration at time t = 3.0s is 18 m/s².