# Quiz Chapter 05.02: Direct Method of Interpolation

 MULTIPLE CHOICE TEST DIRECT INTERPOLATION INTERPOLATION

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

2. 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

3. 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 quadratic polynomial interpolation is most nearly

4. 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) $22$ $24$ $37$ $25$ $123$

$v (t) = 8.667t^{2} - 349.67t + 3523, \, 18 \leq t \leq 24$

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

5. 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

$v(t) = 8.667t^{2} - 349.67t + 3523, \, 18 \leq t \leq 24$

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

6. 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 choose for interpolation are