The position, in feet, of a target is given by s(t)=250-3t+9t^2+2t^3, where t is the time in seconds. The origin of coordinates is at the point where an observer stands, and the positive direction of movement is that one pointing EAST from the origin.
What is the initial position of the target?
What is the initial velocity of the target? (value and direction)
What is the acceleration of the target? Is it constant? Explain.
Find the time/s at which the target isn’t moving.
Find the time/s at which the velocity of the target isn’t changing.
If an observer is moving towards the target according to the position function M(t)=5t+2t^3, find the time when the observer will reach the target.
s(t)=250-3t+9t^2+2t^3
v(t) = -3+18t+6t^2
a(t) = 18+12t
plug in t=0 for initial values.
the other questions are just normal ideas. What do you get?
To answer the questions, we'll need to differentiate the given position function to find the velocity and acceleration functions.
First, let's differentiate s(t) to find the velocity function v(t):
s(t) = 250 - 3t + 9t^2 + 2t^3
Differentiating with respect to time (t), we get:
v(t) = d/dt (250 - 3t + 9t^2 + 2t^3)
= -3 + 18t + 6t^2
To find the acceleration function a(t), we differentiate v(t) with respect to time:
a(t) = d/dt (-3 + 18t + 6t^2)
= 18 + 12t
Now, let's answer the given questions:
1. What is the initial position of the target?
The initial position refers to the position value at t = 0. We can substitute t = 0 into s(t) to find the initial position:
s(0) = 250 - 3(0) + 9(0)^2 + 2(0)^3
= 250
Therefore, the initial position of the target is 250 feet.
2. What is the initial velocity of the target? (value and direction)
The initial velocity refers to the velocity value at t = 0. We can substitute t = 0 into v(t) to get the initial velocity:
v(0) = -3 + 18(0) + 6(0)^2
= -3
The initial velocity of the target is -3 ft/s. The negative sign indicates that the target is initially moving in the opposite direction of the positive direction (east) from the observer.
3. What is the acceleration of the target? Is it constant? Explain.
The acceleration function a(t) is given as a(t) = 18 + 12t. It is not constant because the coefficient of t is not zero. The term 12t indicates that the acceleration changes with time. As time increases, the acceleration will increase linearly.
4. Find the time/s at which the target isn't moving.
To find the time/s when the target is not moving, we need to find the time/s when the velocity v(t) is equal to zero. We can set v(t) = 0 and solve for t:
-3 + 18t + 6t^2 = 0
We can solve this quadratic equation to find the values of t at which the target isn't moving.
5. Find the time/s at which the velocity of the target isn’t changing.
To find the time/s when the velocity is not changing, we need to find the time/s when the acceleration a(t) is equal to zero. We can set a(t) = 0 and solve for t:
18 + 12t = 0
Solving this linear equation will give us the time/s when the velocity isn't changing.
6. If an observer is moving towards the target according to the position function M(t) = 5t + 2t^3, find the time when the observer will reach the target.
To find the time when the observer reaches the target, we need to find the time when the positions of the observer and the target are equal. We can set s(t) = M(t) and solve for t:
250 - 3t + 9t^2 + 2t^3 = 5t + 2t^3
Solving this equation will give us the time when the observer will reach the target.