Question

The position of two particles p1 and the end of t seconds is given by s1=3t^3-12t^2+18t+5, s2=-t^3+9t^2-12t. When do the two particles have the same velocity

Answer 1

To find when the two particles have the same velocity, we need to find when their derivatives are equal. Let's find the derivatives of both position equations:

s1 = 3t^3 - 12t^2 + 18t + 5
s1' = 9t^2 - 24t + 18

s2 = -t^3 + 9t^2 - 12t
s2' = -3t^2 + 18t - 12

Now, let's set the derivatives equal to each other and solve for t:

9t^2 - 24t + 18 = -3t^2 + 18t - 12

Adding 3t^2 - 18t + 12 to both sides:

12t^2 - 6t + 6 = 0

Dividing through by 6:

2t^2 - t + 1 = 0

Using the quadratic formula to solve for t:

t = (-(-1) ± sqrt((-1)^2 - 4(2)(1))) / 2(2)

Simplifying:

t = (1 ± sqrt(1 - 8))/ 4

Since the square root term is -7, which is negative, we can discard the negative solution:

t = (1 + sqrt(1 - 8))/ 4

Simplifying:

t ≈ (1 + sqrt(-7))/4

Since the square root of -7 is an imaginary number, there is no real solution for t. This means that the two particles never have the same velocity.