A particle moves on a line so that its position at time t seconds is p(t)=3t3 meters to the right of the origin. Find the time c in the interval (4,8) for which the velocity of the particle at time c is equal to the average velocity on the interval [4,8].
p(8) = 1536
p(4) = 192
so, avg v is (1536-192)/(8-4) = 336
p'(t) = 9t^2
so, you want
9c^2 = 336
To find the time `c` in the interval (4,8) for which the velocity of the particle at time `c` is equal to the average velocity on the interval [4,8], we need to follow a few steps:
Step 1: Determine the average velocity on the interval [4,8].
- Average velocity is the change in position divided by the change in time. In this case,
average velocity = (change in position) / (change in time)
- The change in position on the interval [4,8] is given by:
Δp = p(8) - p(4) = (3(8)^3) - (3(4)^3) = 1536 - 192 = 1344 meters
- The change in time on the interval [4,8] is 8 - 4 = 4 seconds
- Therefore, the average velocity on the interval [4,8] is:
average velocity = Δp / Δt = 1344 / 4 = 336 meters per second
Step 2: Determine the velocity of the particle at time `c`, where `c` is in the interval (4,8).
- The velocity of the particle at time `t` is given by the derivative of the position function `p(t)`.
- Taking the derivative of p(t), we get:
v(t) = d/dt (3t^3) = 9t^2 meters per second
Step 3: Set up an equation to find the time `c` where v(c) = average velocity.
- Set up the equation: v(c) = average velocity
- Substitute the expressions we derived in steps 1 and 2:
9c^2 = 336
- Divide both sides by 9:
c^2 = 336 / 9 = 37.333
- Take the square root of both sides:
c = ±√(37.333)
Since c is in the interval (4,8), we can ignore the negative square root.
Hence, c = √(37.333) ≈ 6.11
Therefore, the time `c` in the interval (4,8) for which the velocity of the particle at time `c` is equal to the average velocity on the interval [4,8] is approximately 6.11 seconds.