(a) To find the initial velocity, we need to differentiate the equation with respect to time (t). Taking the derivative of v = t³ - 3t² + 2 will give us the velocity function.
The derivative of v with respect to t is dv/dt = 3t² - 6t.
To find the initial velocity, we substitute t = 0 into the derivative:
v(0) = 3(0)² - 6(0)
v(0) = 0 - 0
v(0) = 0
Therefore, the initial velocity is 0.
(b) To find the initial acceleration, we need to differentiate the velocity function with respect to time (t). Taking the derivative of dv/dt = 3t² - 6t will give us the acceleration function.
The derivative of dv/dt with respect to t is d²v/dt² = 6t - 6.
To find the initial acceleration, we substitute t = 0 into the derivative:
a(0) = 6(0) - 6
a(0) = 0 - 6
a(0) = -6
Therefore, the initial acceleration is -6.
(c) To find the acceleration at t = 1 second, we substitute t = 1 into the acceleration function:
Answers & Comments
Answer:
a) 0..... b) -6...... c) 0
Explanation:
(a) To find the initial velocity, we need to differentiate the equation with respect to time (t). Taking the derivative of v = t³ - 3t² + 2 will give us the velocity function.
The derivative of v with respect to t is dv/dt = 3t² - 6t.
To find the initial velocity, we substitute t = 0 into the derivative:
v(0) = 3(0)² - 6(0)
v(0) = 0 - 0
v(0) = 0
Therefore, the initial velocity is 0.
(b) To find the initial acceleration, we need to differentiate the velocity function with respect to time (t). Taking the derivative of dv/dt = 3t² - 6t will give us the acceleration function.
The derivative of dv/dt with respect to t is d²v/dt² = 6t - 6.
To find the initial acceleration, we substitute t = 0 into the derivative:
a(0) = 6(0) - 6
a(0) = 0 - 6
a(0) = -6
Therefore, the initial acceleration is -6.
(c) To find the acceleration at t = 1 second, we substitute t = 1 into the acceleration function:
a(1) = 6(1) - 6
a(1) = 6 - 6
a(1) = 0
Therefore, the acceleration at t = 1 second is 0.