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
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.
To find the time when the two particles have the same velocity, we need to equate their velocities.
The velocity of particle 1 can be found by taking the derivative of its position equation:
v1 = s1' = 9t^2 - 24t + 18.
The velocity of particle 2 can be found by taking the derivative of its position equation:
v2 = s2' = -3t^2 + 18t - 12.
Now, we can set v1 = v2 and solve for t:
9t^2 - 24t + 18 = -3t^2 + 18t - 12.
Combining like terms:
12t^2 - 42t + 30 = 0.
Dividing by 6 to simplify the equation:
2t^2 - 7t + 5 = 0.
Now, we can factor the equation:
(2t - 5)(t - 1) = 0.
Setting each factor equal to zero:
2t - 5 = 0 --OR-- t - 1 = 0.
Solving for t in each case:
2t = 5 --OR-- t = 1.
t = 5/2 --OR-- t = 1.
Therefore, the two particles have the same velocity at t = 5/2 and t = 1.