SOLUTION: An object is moving in a straight line from an intial starting point. Its distance f(x) from starting point at time x is given bt f(x)= f(x)=2x4 + 3x3 - 1x2 + 3x

SOLUTION: An object is moving in a straight line from an intial starting point. Its distance f(x) from starting point at time x is given bt f(x)= f(x)=2x4 + 3x3 - 1x2 + 3x - 1 the velocity

Question 63934: An object is moving in a straight line from an intial starting point. Its distance f(x) from starting point at time x is given bt f(x)=
f(x)=2x4 + 3x3 - 1x2 + 3x - 1
the velocity of the object is given by f(x) and the acceleration of the object is given by f(x)
A) what is the distance from the starting point when x = 10.0?
B) what is the velocity of the object when x=4.0?
C) what is the acceleration of the object when x = 11.6??
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
An object is moving in a straight line from an intial starting point. Its distance f(x) from starting point at time x is given bt f(x)=
f(x)=2x^4 + 3x^3 - x^2 + 3x - 1
the velocity of the object is given by f(x) and the acceleration of the object is given by f(x)
A) what is the distance from the starting point when x = 10.0?
Find f(10)
B) what is the velocity of the object when x=4.0?
Velocity = 1st derivative = f'(x)=8x^3+x^2-2x+3
Find f'(4)

C) what is the acceleration of the object when x = 11.6??
Acceleration = 2nd derivative = f''(x)=24x^2+2x-2
Find f''(11.6)
--------
Cheers,
Stan H.
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