Calculus natural science

Question

A particle moves according to a law of motion s = f(t) = t3 - 12t2 + 45t, t 0, where t is measured in seconds and s in feet.

a.When is the particle speeding up? (????,????)U(????,????)

b.When is it slowing down?

(????,????)U(????,????)

Answer 1

Given

Function of Position: s(t) = t³ - 12t²+45t

Find

a) When particle speeds up

b) When particle slows down

Solution

s(t) = t³ - 12t²+45t

Please note to find the velocity function we take the derivative

We use the power rule to derive this polynomial.

v(t) = 3t² - 24t+45

If we plot this velocity function and identify the vertex (axis of symmetry).

Vertex = -b/2a = 4

The particle will slow down when the time is less than 4 seconds, but will speed up when

it is greater than 4 second. We can see this graphically by plotting the velocity function.

If we approach the function from the left, you see the velocity decreasing until it reaches the vertex.

Once past the vertex the velocity increases.

A more intuitive way is to take the derivative of the velocity function.

a = 6t -24

The acceleration by definition is the rate of change of velocity.

Acceleration function is negative when it is less then 4, meaning velocity is decreasing at a certain rate.

Acceleration function is positive when it is greater then 4, meaning velocity is increasing at a certain rate.

In summary,

Particle speeds up when t>4

Particle slows down when t<4

Answer 2

By definition the velocity v(t) is the derivative of s(t). Hence, v(t)=3t^2-24t+45. Since we want to find when the particle is accelerating it means that we are interested in knowing when the function v(t) is increasing. To answer this we calculate the acceleration a(t)=v'(t)=6t-24 and solve the inequality a(t)>=0 which is satisfied for t>=4 seconds. Thus, the particle is accelerating in the interval [4,\infty).