Question

The position of a particle moving along the x-axis varies with time according to x(t)=5.0t²−4.0t³ m. Find the velocity and acceleration of the particle as functions of time.

a) v(t)=10.0t−12.0t² ,a(t)=10.0−24.0t
b) v(t)=10.0t−12.0t² ,a(t)=−24.0t
c) v(t)=10.0−12.0t,a(t)=10.0−24.0t
d) v(t)=10.0−12.0t,a(t)=−24.0t

Answer 1

The velocity of the particle is v(t)=10.0t-12.0t² m/s and the acceleration is a(t)=10.0-24.0t m/s².

Step-by-step explanation:

To find the velocity and acceleration as functions of time for a particle whose position varies according to the equation x(t) = 5.0t² - 4.0t³, we need to differentiate the position function with respect to time.

Velocity Function

The velocity v(t) is the first derivative of the position function x(t). Differentiating x(t) with respect to time gives us:

v(t) = d/dt [5.0t² - 4.0t³] = 10.0t - 12.0t² m/s.

Acceleration Function

The acceleration a(t) is the derivative of the velocity function. So, we differentiate v(t) to get:

a(t) = d/dt [10.0t - 12.0t²] = 10.0 - 24.0t m/s².

Hence, the correct answer is option a): v(t)=10.0t−12.0t², a(t)=10.0−24.0t.