The distance meters from a fixed point O, covered by a particle after t seconds given by the equation; S = t3 – 6t2 + 9t + 5.

(a) Calculate the gradient to the curve at t = 0.5 seconds
(b) Determine the values of sat the maximum and minimum turning points of the curve

dS/dt = 3t^2-12t+9

for max/min, find t when dS/dt = 0

To find the gradient to the curve at t = 0.5 seconds, we need to calculate the derivative of the equation S with respect to t and substitute t = 0.5.

(a) Calculate the gradient to the curve at t = 0.5 seconds:

Step 1: Find the derivative of the equation S with respect to t.
dS/dt = 3t^2 - 12t + 9

Step 2: Substitute t = 0.5 into the derivative equation to find the gradient at t = 0.5 seconds.
dS/dt = 3(0.5)^2 - 12(0.5) + 9
= 1.5 - 6 + 9
= 4.5

Therefore, the gradient to the curve at t = 0.5 seconds is 4.5.

(b) Determine the values of s at the maximum and minimum turning points of the curve:

Step 1: Find the second derivative of the equation S with respect to t.
d^2S/dt^2 = 6t - 12

Step 2: Set the second derivative equation equal to zero and solve for t to find the values of t at the maximum and minimum turning points.

6t - 12 = 0
6t = 12
t = 2

So, t = 2 is the value of t at both the maximum and minimum turning points.

Step 3: Substitute t = 2 into the original equation S to find the corresponding values of s at the maximum and minimum turning points.

S = t^3 – 6t^2 + 9t + 5
S = (2)^3 – 6(2)^2 + 9(2) + 5
S = 8 - 24 + 18 + 5
S = 7

Therefore, at both the maximum and minimum turning points, the value of s is 7.

To calculate the gradient to the curve at t = 0.5 seconds, we need to find the derivative of the equation with respect to time (t) and then substitute t = 0.5 into the derivative equation.

(a) Calculate the gradient at t = 0.5 seconds:
To find the derivative of the equation, we differentiate each term of the equation with respect to t. Here's the step-by-step process:

1. Differentiating t^3:
The derivative of t^3 with respect to t is 3t^2.

2. Differentiating -6t^2:
The derivative of -6t^2 with respect to t is -12t.

3. Differentiating 9t:
The derivative of 9t with respect to t is 9.

4. Differentiating 5:
The derivative of a constant term (in this case, 5) is 0.

Now, we combine these derivatives to get the derivative equation, which represents the gradient at any value of t:
S' = 3t^2 - 12t + 9

To calculate the gradient at t = 0.5 seconds, substitute t = 0.5 into the derivative equation:
S'(0.5) = 3(0.5)^2 - 12(0.5) + 9
= 3(0.25) - 6 + 9
= 0.75 - 6 + 9
= 3.75

Therefore, the gradient to the curve at t = 0.5 seconds is 3.75.

(b) To determine the values of s at the maximum and minimum turning points of the curve, we need to find the values of t at which the derivative of the equation is equal to zero. These values of t will correspond to the maximum and minimum points.

To find these values, we set the derivative equation S' equal to zero and solve for t:

3t^2 - 12t + 9 = 0

We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 3, b = -12, and c = 9.

t = (-(-12) ± √((-12)^2 - 4(3)(9))) / (2(3))
t = (12 ± √(144 - 108)) / 6
t = (12 ± √36) / 6
t = (12 ± 6) / 6

This gives us two possible values for t:

1. t = (12 + 6) / 6 = 18 / 6 = 3
2. t = (12 - 6) / 6 = 6 / 6 = 1

Thus, the values of t corresponding to the maximum and minimum turning points of the curve are t = 1 and t = 3.

To find the corresponding values of s at these points, substitute these values back into the equation S = t^3 - 6t^2 + 9t + 5:

1. s = (1)^3 - 6(1)^2 + 9(1) + 5 = 1 - 6 + 9 + 5 = 9
2. s = (3)^3 - 6(3)^2 + 9(3) + 5 = 27 - 54 + 27 + 5 = 5

Therefore, the values of s at the maximum and minimum turning points of the curve are s = 9 and s = 5, respectively.