The distance is given by
y = 58t - 1.86t²
The velocity is the derivative of y with respect to t. It is
v = 58 - 3.72t
In an interval [t₁, t₂], the average velocity is
![\bar{v} = \frac{1}{t_{2}-t_{1}}[58t-1.86t^{2}]_{t1}^{t2}](https://tex.z-dn.net/?f=%5Cbar%7Bv%7D%20%3D%20%5Cfrac%7B1%7D%7Bt_%7B2%7D-t_%7B1%7D%7D%5B58t-1.86t%5E%7B2%7D%5D_%7Bt1%7D%5E%7Bt2%7D)
Evaluate the results (Use v to denote the average velocity).
i. t = [1,2]
v = (108.56 - 56.14)/1 = 52.42
Answer: 52.42 m/s
ii. t=[1,1.5]
v = (82.815 - 56.14)/0.5 = 53.35
Answer: 53.35 m/s
iii. t=[1,1.1]
v = (61.5494 - 56.14)/0.1 = 54.094
Answer: 54.094 m/s
iv. t=[1,1.01]
v = (56.6826 - 56.14)/0.01 = 54.26
Answer: 54.26 m/s
Note:
As the time interval approaches zero, we obtain the derivative of y at t=1, which is 58-3.72 = 54.28 m/s
y = 58t - 1.86t²
The velocity is the derivative of y with respect to t. It is
v = 58 - 3.72t
In an interval [t₁, t₂], the average velocity is
Evaluate the results (Use v to denote the average velocity).
i. t = [1,2]
v = (108.56 - 56.14)/1 = 52.42
Answer: 52.42 m/s
ii. t=[1,1.5]
v = (82.815 - 56.14)/0.5 = 53.35
Answer: 53.35 m/s
iii. t=[1,1.1]
v = (61.5494 - 56.14)/0.1 = 54.094
Answer: 54.094 m/s
iv. t=[1,1.01]
v = (56.6826 - 56.14)/0.01 = 54.26
Answer: 54.26 m/s
Note:
As the time interval approaches zero, we obtain the derivative of y at t=1, which is 58-3.72 = 54.28 m/s