Question

S(t)=-t^3+2t^2+(3/2) be the position of a particle moving along the x-axis at time t. At what time will the instantaneous velocity equal the average velocity over the time interval (0,4)?

Answer 1

(2 + 2sqrt(7))/3

Approx 2.43

Explanation:

The average velocity over [0,4] is

(s(4)-s(0))/(4-0)

So we need to find s(4) and s(0).

s(t)=-t^3+2t^2+(3/2)

s(4)=-4^3+2(4)^2+(3/2)

s(4)=-64+32+(3/2)

s(4)=-32+(3/2)

s(4)=-61/2

s(t)=-t^3+2t^2+(3/2)

s(0)=-0^3+2(0)^2+(3/2)

s(0)=3/2

s(4)-s(0)=-61/2-3/2=-64/2=-32

So the average velocity over the given interval is -32/4=-8.

Now we wanr to find t such that the instantaneous velocity, s'(t), is equal to the average velocity on the given interval

So we want to solve s'(t)=-8.

Let's differentiate.

s(t)=-t^3+2t^2+(3/2)

s'(t)=-3t^2+4t+0

We will need to solve the quadratic equation

-3t^2+4t=-8

Multiply both sides by -1

3t^2-4t=8

Subtract 8 on both sides

3t^2-4t-8=0

The discriminant,D, is b^2-4ac=(-4)^2-4(3)(-8)=16-12(-8)=16+96=112.

The quadratic formula is (-b pm sqrt(D))/(2a) where pm means plus or minus.

(4 pm sqrt(112))/6

sqrt(112)

sqrt(16)sqrt(7)

4sqrt(7)

So the solutions from the quadratic equation can be simplified further

(4 pm sqrt(112))/6

(4 pm 4sqrt(7))/6

(2 pm 2sqrt(7))/3