Kinematics Basics: Solving a & b with a=-8m/sec2

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The discussion focuses on solving kinematics problems involving a particle with a constant acceleration of -8 m/sec². The user seeks to find the time when the velocity is zero and the velocity and total distance traveled at t = 11 seconds. It is suggested to use the equation s(t) = (1/2) a t² + C t + D to determine the constants C and D, which represent initial velocity and displacement, respectively. By substituting the known values into this equation, the user can derive the necessary parameters to solve both parts of the problem. The conversation emphasizes the importance of understanding the relationship between acceleration, velocity, and displacement in kinematics.
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Homework Statement



The acceleration of a particle is defined by a = -8 m/sec2. Knowing that x = 20 when t = 4 and x = 4 when v = 16, determine:

  • The time when the velocity is zero
  • The velocity and the total distance traveled when t = 11

Homework Equations





The Attempt at a Solution



I think once I get part 'a' I can do part 'b'. Here is my attempt:

a = -8m/sec2
So
V = -8t + C.

How can I get C? Do I need to use another representation of acceleration? I am not sure where to go.

Any advice greatly appreciated!
Adrian
 
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C is the velocity at time t = 0.
I would suggest using the formula
s(t) = (1/2) a t² + C t + D (*)
where C is the velocity v(0) at time t = 0 and D is the displacement s(0) at t = 0.

If you plug in the two given points you can determine C and D. Then you can use both formula (note that v(t) = -8t + C will actually be the derivative of formula (*) above).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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