This acceleration function represents the acceleration of a subatomic particle:
a(t) = 6t² - 8t
The particle has an initial velocity of 1 m/s. What is the velocity of the particle at t = 2 s?



Answer :

Answer: The velocity of the particle at time = 2s is 65 m/s.

Explanation:

Velocity Equation

Using the provided acceleration function, we can work backward to find the particle's velocity function.

Acceleration is the derivative of velocity (with respect to time), so we can take the antiderivative of the a(t) function to get the v(t) function. To take an antiderivative, multiply the coefficient of each variable by its exponent. Each exponent will increase by one as well. Don't forget to add "+ C" at the end to represent some constant that we don't know the value of.

a(t) = 6t² - 8t

v(t) = (6*2)t³ - (8*1)t² + C

v(t) = 12t³ - 8t² + C

In the problem, we're given v(0), or initial velocity, is equal to 1 m/s. We can set our velocity function equal to v(0) to find C and complete the equation.

v(0) = 1 = 12(0)³ - 8(0)² + C

1 = 0 - 0 + C

1 = C

So, our velocity function is v(t) = 12t³ - 8t² + 1.

Solving for t = 2

Now for the easy part, we can plug 2 in for t in our velocity function to find the particle's velocity at t = 2.

v(t) = 12t³ - 8t² + 1

v(2) = 12(2)³ - 8(2)² + 1

v(2) = 12(8) - 8(4) + 1

v(2) = 96 - 32 + 1

v(2) = 65 m/s