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 MULTIPLE CHOICE TEST DIRECT INTERPOLATION INTERPOLATION

Q1. Given n+1 data pairs, a unique polynomial of degree ________________ passes through the n+1 data points
n+1

n+1 or less
n
n or less

Q2. The following data of the velocity of a body is given as a function of time.

 Time (s) 0 15 18 22 24 Velocity (m/s) 22 24 37 25 123

The velocity in m/s at 16 s using linear polynomial interpolation is most nearly

27.867
28.333
30.429
43.000

Q3. The following data of the velocity of a body is given as a function of time.

 Time (s) 0 15 18 22 24 Velocity (m/s) 22 24 37 25 123

The velocity in m/s at t=16 s using quadratic polynomial interpolation is most nearly

27.867
28.333
30.429
43.000

Q4. The following data of the velocity of a body is given as a function of time

 Time (s) 0 15 18 22 24 Velocity (m/s) 22 24 37 25 123

approximates the velocity of the body between 18 and 24 seconds.  From this information, one of the times at which the velocity of the body is 35 m/s during the above time interval of t=18 s to 24 s is

18.67 s
20.85 s

22.20 s
22.29 s

Q5. The following data of the velocity of a body is given as a function of time.

 Time (s) 0 15 18 22 24 Velocity (m/s) 22 24 37 25 123

One of the interpolant approximations from the above data is given as

Using the above interpolant, the distance in meters covered by the body between t=19 s and t=22 s is most nearly

10.50

88.50

93.00

168.00

Q6. The following data of the velocity of a body is given as a function of time

 Time (s) 0 15 18 22 24 Velocity (m/s) 22 24 37 25 123

If you were going to use quadratic interpolation to find the value of the velocity at t=14.9 seconds, the three data points of time that you would you choose for interpolation are

0, 15, 18
15,18,22
0, 15, 22
0, 18, 24

Complete Solution

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