Can you answer this calculus question?

Question

A particle moves on the x-axis so that its velocity at any time t≥0 is given by v(t)=12t^2-36t+15. At t=1, the particle is at the origin. 

 

a) Find the maximum velocity of the particle for 0≤t≤2. 

b) Find the total distance traveled by the particle from t=0 to t=2. 

 

Please show all your work. 

Answer 1

A particle moves on the x-axis so that its velocity at any time t≥0 is given by v(t)=12t^2-36t+15. At t=1, the particle is at the origin.

a) Find the maximum velocity of the particle for 0≤t≤2.

 

a(t) = v'(t) = 24t-36 = 0 when t = 3/2
v(t=3/2) = -12

 

This is the minimum.  Since coefficient of t² term is positive (12), parabola opens upward with vertex at t = 3/2.  Max value will be at t=0 or t = 2:

 

v(t=0) = 15  (max)

v(t=2) = -9

 

b) Find the total distance traveled by the particle from t=0 to t=2.

 

          2                    2                 2               2

d(t) = ∫v(t)dt =  12∫t² dt - 36∫t dt + 15∫dt

         0                    0                 0                0

 

d(t) = (4t³ - 18t² + 15t)  evaluate at d(t=2) - d(t=0)

Answer 2

A particle moves on the x-axis so that its velocity at any time t≥0 is given by v(t) = 12t^2 - 36t + 15. At t=1, the particle is at the origin.

a) Find the maximum velocity of the particle for 0≤t≤2.

v(t) is a parabola that opens up because the leading coefficient is positive. So the vertex is a minimum, not a maximum. Let’s find it, though:
h = 36/24 = 3/2, k = 15 - 12(3/2)^2 = 15-27 = -12
Vertex = (3/2,-12)

The maximum on the interval 0≤t≤2 must occur at one of the endpoints:

v(0) = 15
v(2) = 12(2)^2 - 36(2) + 15 = 48-72+15 = 15-24 = -9

So the maximum velocity of the particle for 0≤t≤2 is +15.

b) Find the total distance traveled by the particle from t=0 to t=2.

v(t) = 12t^2 - 36t + 15

Zeros of v:

12t^2 - 36t + 15 = 0

4t^2 - 12t + 5 = 0

(2t-1)(2t-5) = 0

t = 1/2 or t = 5/2 > 2

v(t) > 0 for 0 < t < 1/2
v(t) < 0 for 1/2 < t < 2

Total distance traveled, D, is:

D = ∫{0,1/2}(v(t)dt) – ∫{1/2,2}(v(t)dt)

= [4t^3-18t^2+15t]{0,1/2}-[4t^3-18t^2+15t]{1/2,2}

= 2[4(1/2)^3-18(1/2)^2+15(1/2)]-[4(2)^3-18(2)^2+15(2)]

= 2[1/2-9/2+15/2]-[32-72+30]

= 7-[-10]

= 17