Find the maximum velocity on the interval [0,4] if a(t)=6t-12 for 0≤ t ≤4 and if v(0)=18.?

Answer 1

Since the acceleration function a(t) is the instantaneous slope of the velocity and the relative max/min occurs where the slope is 0 ...

Set a(t) = 0 to find the value of t when the max/min velocity was reached.
0 = 6t -12
t=2

2 is in the interval [0,4], so it can be used, BUT t=2 is a MINIMUM. (To the left of t=2, a(t) is negative -- showing deceleration, the right of t=2 a(t) is positive -- showing acceleration.)

Find the velocity function by integrating a(t).
v(t) = 3t^2 - 12t + c

To find c, plug in the point (0, 18).

The velocity function is a parabola opening up. With t=2 as the minimum in the interval [0.4], plug in the end points (t = 0 or 4) to see which is the greatest velocity.

Answer 2

a(t) = dv/dt = 6t - 12
Integrating v = 3(t^2) - 12t + c
Given at: t = 0, v = 18
So, c = 18
Hence, v = 3t^2 - 12t + 18 = 3(t^2 - 4t + 6) = 3[(t - 2)^2 + 2]

So v is maximum when t is maximum.
In the interval [0,4] t is max when t = 4
So, v = 3[(4 - 2)^2 + 2] = 18

At t = 0 also i.e. at start velocity is maximum and the value is 18.

Answer 3

v(t) = 3t² - 12t + c
v(0) = 18
c = 18
v(t) = 3t² - 12t + 18
dv/dt = 6t - 12 = 0 for turning point on graph
t = 2 at turning point
v(2) = 12 - 24 + 18
v(2) = 6
(2,6) is a turning point
d²v/dt² = 6 is + ve
Thus (2,6) is a Minimum turning point
v(0) = 18
v(4) = 18
v(2) = 6

Maximum velocity on given interval = 18

Answer 4

V(0)=18, V is at it's Max when time is equal zero

Answer 5

Find it yourself...
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