these are two similar kinematics maths question which i don't know. Please help!
1)A particle moves in a straight line so that its distance, s metres, from a fixed point O is given by s = 2t^3 - 3t^2 - 12t + 6, where t is the time in seconds after passing O. Find the minimum speed attained by the particle
2)A particle moves in a straight line so that its distance, s meters, from a fixed point is given by s=t^3(2-t)^2, where t is the time in seconds after passing through the fixed point. What is the greatest distance travelled by the particle?
Bailey, Andrea, Danny, Riley, anabelle, holly, LILLY, zachary, I'm stumped -- or whoever!
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1) The speed at time t is
|ds/dt| = |6t^2 -6t -12|
(Speed is always positive)
The minimum value of |ds/dt| is 0 if there is a time when
6t^2 -6t -12 = 0
t^2 -t -2 = 0
(t-2)(t+1) = 0
So there IS a time when the speed is zero.
2) The greatest value of s occurs at a time where ds/dt = 0 is the second derivative d^2/dt^2 at that time is negative.
s grows without bound for very large t greater than 2. I do not see how there could be a greatest value of s unless you restrict the region of consideration to 0 < t <2
t = 2
In my answer to 2), "is" should be "if".
Sorry about that
Sure, let's solve these two kinematics questions step by step:
1) To find the minimum speed attained by the particle, we need to determine when the speed is at its lowest. In this case, speed refers to the rate of change of distance with respect to time, which is the derivative of the distance function.
Given that the particle's distance from the fixed point O is given by s = 2t^3 - 3t^2 - 12t + 6, we can find its speed by differentiating s with respect to time (t). So, let's find ds/dt:
ds/dt = d/dt (2t^3 - 3t^2 - 12t + 6)
= 6t^2 - 6t - 12
Now we have the expression for the speed of the particle. To find the minimum speed, we need to determine the critical points of this function. The critical points occur when the derivative is equal to zero:
6t^2 - 6t - 12 = 0
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-(-6) ± sqrt((-6)^2 - 4(6)(-12))) / (2(6))
t = (6 ± sqrt(36 + 288)) / 12
t = (6 ± sqrt(324)) / 12
t = (6 ± 18) / 12
Thus, we have two possible solutions for t: t1 = 4/3 and t2 = -2.
Now, we need to determine which value of t gives us the minimum speed. We can do this by substituting these values into the speed equation:
For t = 4/3:
Speed at t1 = 6(4/3)^2 - 6(4/3) - 12
= 32/3 - 24/3 - 12
= -20/3
For t = -2:
Speed at t2 = 6(-2)^2 - 6(-2) - 12
= 24 + 12 - 12
= 24
Therefore, the minimum speed attained by the particle is -20/3 units.
2) To find the greatest distance traveled by the particle, we need to determine the maximum value of the distance function. In this case, the distance is given by s = t^3(2 - t)^2.
To find the maximum distance, we can take the derivative of the distance function with respect to time (t) and set it equal to zero. Let's find ds/dt:
ds/dt = d/dt [t^3(2 - t)^2]
= 3t^2(2 - t)^2 + t^3(2)(2 - t)(-1)
To find the critical points, we set the derivative equal to zero:
3t^2(2 - t)^2 + t^3(2)(2 - t)(-1) = 0
We can simplify this equation and solve for t. However, instead of doing that, let's look for potential critical points by factoring:
3t^2(2 - t)^2 + t^3(2)(2 - t)(-1) = 0
t^2(2 - t)[3(2 - t) - 2t] = 0
t^2(2 - t)(6 - 5t) = 0
From this equation, we can see that there are three possible values for t that could give us critical points: t1 = 0, t2 = 2, and t3 = 6/5.
Now, we need to determine which value of t gives us the maximum distance. We can do this by plugging each value into the distance equation and comparing the results:
For t = 0:
Distance at t1 = (0)^3(2 - 0)^2 = 0
For t = 2:
Distance at t2 = (2)^3(2 - 2)^2 = 0
For t = 6/5:
Distance at t3 = (6/5)^3(2 - 6/5)^2 = (216/125)(4/5)^2 = (216/125)(16/25) = 3456/625
Therefore, the greatest distance traveled by the particle is 3456/625 units.
I hope this helps! Let me know if you have any further questions.