The height of a particle is given by the equation f(t)=t^2+2t+6
(a) What is the average velocity of the particle on the interval [2,3]?b.What is its instantaneous velocity at t = 2.
What is the instantaneous velocity of the object at any time t?
avg velocity= distance/time
= (height(t=3)-height(t=2)/(3-2)
instantaneous velocity? df/dt=2t+2
To find the average velocity of the particle on the interval [2, 3], you need to calculate the change in height divided by the change in time.
(a) Average velocity on the interval [2, 3]:
1. Find the height at t = 2: Substitute t = 2 into the equation f(t) = t^2 + 2t + 6.
f(2) = (2)^2 + 2(2) + 6
= 4 + 4 +6
= 14
2. Find the height at t = 3: Substitute t = 3 into the equation f(t) = t^2 + 2t + 6.
f(3) = (3)^2 + 2(3) + 6
= 9 + 6 + 6
= 21
3. Calculate the change in height: The change in height is f(3) - f(2).
Change in height = 21 - 14
= 7
4. Calculate the change in time: The change in time is 3 - 2.
Change in time = 3 - 2
= 1
5. Calculate the average velocity: Divide the change in height by the change in time.
Average velocity = (Change in height) / (Change in time)
= 7 / 1
= 7
Therefore, the average velocity of the particle on the interval [2,3] is 7.
(b) To find the instantaneous velocity at t = 2, you need to find the derivative of the equation f(t) = t^2 + 2t + 6.
1. Find the derivative of f(t) with respect to t: Differentiate each term of f(t) separately.
f'(t) = 2t + 2
2. Substitute t = 2 into the derived equation to find the instantaneous velocity.
f'(2) = 2(2) + 2
= 4 + 2
= 6
Therefore, the instantaneous velocity of the particle at t = 2 is 6.
The instantaneous velocity of the object at any time t can be found by evaluating the derivative of the given equation f(t). The derivative represents the rate of change of the height with respect to time, which gives the instantaneous velocity.