Asked by Jane
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.
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.
Answers
Answered by
Steve
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?
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?
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