The 58 m/s initial velocity of the arrow with height, y = 58•t - 0.83•t², gives;
(1) 55.925 m/s
(2) 56.257 m/s
(3) 56.3317 m/s
(4) 56.33917 m/s
(b) 56.34 m/s
How can the average and instantaneous velocity be calculated?
Initial velocity of the arrow = 58 m/s
Height of the arrow is; y = 58•t - 0.83•t²
The above kinematic equation can be analyzed as follows;
(1) The height at t = 1 is found as follows;
y = 58 × 1 - 0.83 × 1² = 57.17
Height at t = 1.5 s. is y = 58×1.5 - 0.83×1.5² = 85.1325
Which gives;
Average velocity = (85.1325-57.17)/(1.5-1) = 55.925
The average velocity in the interval [1, 1.5] is therefore;
(2) [1, 1.1]
Height at t = 1.1 s. is y = 58×1.1 - 0.83×1.1² = 62.7957
Which gives;
Average velocity = (62.7957-57.17)/(1.1-1) = 56.257
- The average velocity is 56.257 m/s
(3) [1, 1.01]
Height at t = 1.01 s. is y = 58×1.01 - 0.83×1.01² = 57.733317
Which gives;
Average velocity = (57.733317-57.17)/(1.01-1) = 56.3317
The average velocity is 56.33 m/s
(4) [1, 1.001]
Height at t = 1.01 s. is y = 58×1.001 - 0.83×1.001² = 57.22633917
Which gives;
Average velocity = (57.22633917-57.17)/(1.001-1) = 56.33917
- The average velocity is 56.33917 m/s
(b) The equation for the instantaneous velocity is found as follows;
v = dy/dt = 58 - 2×0.83•t
After 1 second, we have;
t = 1
v = 58 - 2×0.83 × 1 = 56.34
- The instantaneous velocity at t = 1 seconds is 56.34 m/s
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