The position of a point on a line is given by the equation s(t)= 2t^3-3t^2-30t+36 where s is measured in metres and t in seconds.
a) what is the velocity of the point after 2 seconds?
b) what is its acceleration after 3 seconds?
c) where is it when it first stops moving?
d) After 6 seconds, is it moving toward or moving away from the origin?
per the usual rules,
(a) v(t) = 6t^2-6t-30 = 6(t^2-t-5)
(b) a(t) = 6(2t-1)
(c) solve for v=0 and then evaluate s there
(d) is v(6) positive or negative?
a) Well, if you want to know the velocity after 2 seconds, we need to find the derivative of the equation s(t). Let's do some math magic: v(t) = s'(t) = 6t^2 - 6t - 30. Now, plug in t = 2 and calculate v(2). Let me grab my calculator... oops, I forgot to bring it. But I can tell you that v(2) will give you the velocity at 2 seconds.
b) To find the acceleration after 3 seconds, we need to differentiate the velocity function we just found. So, a(t) = v'(t) = 12t - 6. Plug in t = 3, and voila! You'll have the acceleration after 3 seconds. Stay safe from all that acceleration!
c) Ah, the moment when it stops moving! To find the position at that moment, we need to set the velocity equation, 6t^2 - 6t - 30, equal to zero and solve for t. Once we have that value of t, we substitute it into the position equation s(t) = 2t^3 - 3t^2 - 30t + 36. It might take some time, but when we find it, we'll know where the point first decided to take a break.
d) After 6 seconds, we need to check if the point is moving toward or away from the origin. We can find this out by examining the sign of the velocity at t = 6. If v(6) is positive, it's moving away from the origin. If it's negative, the point is coming back closer to the origin. So, grab your calculator again and let's plug in t = 6 to determine the direction of its travel.
a) To find the velocity of the point after 2 seconds, we need to calculate the derivative of the position equation, s(t), with respect to time, t.
The derivative of s(t) with respect to t is given by:
v(t) = d/dt [2t^3 - 3t^2 - 30t + 36]
To differentiate the equation, we use the power rule for derivatives. Differentiating term by term, we get:
v(t) = 6t^2 - 6t - 30
Now, substitute t = 2 into the velocity equation:
v(2) = 6(2)^2 - 6(2) - 30
= 6(4) - 12 - 30
= 24 - 12 - 30
= -18
Therefore, the velocity of the point after 2 seconds is -18 m/s.
b) To find the acceleration of the point after 3 seconds, we need to calculate the derivative of the velocity equation, v(t), with respect to time, t.
The derivative of v(t) with respect to t is given by:
a(t) = d/dt [6t^2 - 6t - 30]
Again, using the power rule for derivatives, we differentiate each term:
a(t) = 12t - 6
Substitute t = 3 into the acceleration equation:
a(3) = 12(3) - 6
= 36 - 6
= 30
Therefore, the acceleration of the point after 3 seconds is 30 m/s^2.
c) To find where the point first stops moving, we need to find the values of t that make the velocity, v(t), equal to zero.
In the velocity equation, v(t) = 6t^2 - 6t - 30, set v(t) = 0 and solve for t:
6t^2 - 6t - 30 = 0
Divide the equation by 6 to simplify:
t^2 - t - 5 = 0
This quadratic equation can be factored as follows:
(t - 2)(t + 5) = 0
Setting each factor equal to zero, we find two possible values for t:
t - 2 = 0 --> t = 2
t + 5 = 0 --> t = -5
However, t represents time, so we discard the negative value, t = -5.
Therefore, the point first stops moving at t = 2 seconds.
d) To determine if the point is moving toward or away from the origin after 6 seconds, we need to analyze the sign of the velocity, v(t).
Using the given velocity equation, v(t) = 6t^2 - 6t - 30, substitute t = 6:
v(6) = 6(6)^2 - 6(6) - 30
= 6(36) - 36 - 30
= 216 - 36 - 30
= 150
Since the velocity is 150 m/s, which is positive, the point is moving away from the origin after 6 seconds.
To find the velocity, acceleration, and the position at a certain time, we need to differentiate the given equation. Let's go through each question step by step:
a) Velocity after 2 seconds:
To find the velocity, we need to differentiate the given equation with respect to time (t). Differentiating the equation s(t) = 2t^3 - 3t^2 - 30t + 36 will give us the equation for velocity.
Here's how you can do it:
1. Differentiate each term separately using the power rule:
- Differentiating 2t^3 gives 6t^2
- Differentiating -3t^2 gives -6t
- Differentiating -30t gives -30
- Differentiating 36 gives 0
2. Combining all the differentiated terms, we get:
v(t) = 6t^2 - 6t - 30
Now, to find the velocity after 2 seconds, substitute t = 2 into the velocity equation:
v(2) = 6(2)^2 - 6(2) - 30
v(2) = 24 - 12 - 30
v(2) = -18
Therefore, the velocity of the point after 2 seconds is -18 m/s.
b) Acceleration after 3 seconds:
To find the acceleration, we need to differentiate the velocity equation (v(t) = 6t^2 - 6t - 30) with respect to time (t).
Here's how you can do it:
1. Differentiate each term separately using the power rule:
- Differentiating 6t^2 gives 12t
- Differentiating -6t gives -6
- Differentiating -30 gives 0
2. Combining all the differentiated terms, we get:
a(t) = 12t - 6
Now, to find the acceleration after 3 seconds, substitute t = 3 into the acceleration equation:
a(3) = 12(3) - 6
a(3) = 36 - 6
a(3) = 30
Therefore, the acceleration of the point after 3 seconds is 30 m/s^2.
c) Position when it first stops moving:
When the point stops moving, the velocity becomes zero. To find the position when it first stops moving, we need to find the value of t when v(t) = 0.
Using the velocity equation v(t) = 6t^2 - 6t - 30, we set it equal to zero and solve for t:
6t^2 - 6t - 30 = 0
To solve this quadratic equation, you can factor it, complete the square, or use the quadratic formula. Let's use the quadratic formula:
t = (-(-6) ± √((-6)^2 - 4(6)(-30)))/(2(6))
t = (6 ± √(36 + 720))/(12)
t = (6 ± √756)/(12)
t = (6 ± 2√189)/(12)
t = (1 ± √189)/2
Since time (t) cannot be negative, we only consider the positive value:
t ≈ (1 + √189)/2 ≈ 6.066
Therefore, the point first stops moving at approximately 6.066 seconds.
d) Movement towards or away from the origin after 6 seconds:
To determine whether the point is moving toward or away from the origin after 6 seconds, we need to check the sign of the velocity at that time.
Substituting t = 6 into the velocity equation v(t) = 6t^2 - 6t - 30:
v(6) = 6(6)^2 - 6(6) - 30
v(6) = 216 - 36 - 30
v(6) = 150
Since the velocity is positive at t = 6 (v(6) = 150 > 0), the point is moving away from the origin after 6 seconds.
Note: In this context, positive velocity indicates movement in the positive direction, while negative velocity represents movement in the negative direction.