SOLUTION: Please help me solve this problem: An object is moving in a straight line from a fixed point. The displacement s (in meters) is given by s= -2t^2 + 28t + 45, t≥0 , where

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Question 1006614: Please help me solve this problem:
An object is moving in a straight line from a fixed point. The displacement s (in meters) is given by s= -2t^2 + 28t + 45, t≥0 , where t is in seconds.
a) Find the velocity at any time.
b) What is the velocity at t=3s?
c) Find t when the object reaches its maximum displacement.
d) Find the maximum displacement reached by this object.
e) Determine the acceleration at any time.
d) Describe the direction of the motion at, t=3s, t=7s, and t=9s.

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
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An object is moving in a straight line from a fixed point. The displacement s (in meters) is given by s= -2t^2 + 28t + 45, t≥0 , where t is in seconds.
a) Find the velocity at any time.
b) What is the velocity at t=3s?
c) Find t when the object reaches its maximum displacement.
d) Find the maximum displacement reached by this object.
e) Determine the acceleration at any time.
d) Describe the direction of the motion at, t=3s, t=7s, and t=9s.
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a) Velocity is the derivative %28ds%29%2F%28dt%29 = -4t + 28 m%2Fs.

b) Velocity at t=3s is v = (-4)*3 + 28 = -12 + 28 = 16 m%2Fs.

c) It is the time moment when the speed is zero: -4t + 28 = 0 -----> t = 28%2F4 = 7 sec. It is the same time moment when %28ds%29%2F%28dt%29 = 0.

d) The maximum displacement reached by this object is s at t = 7 sec. Substitute t = 7 into s= -2t^2 + 28t + 45 to get s = 143 m.

e) Acceleration is the second derivative of s or the derivative of the velocity a = -2 m%2Fs%5E2.

f) At t=3s the body moves to "positive s". At t=7s the speed is zero. At t=9s the body moves to "negative s".