Answer:
The answer is below
Explanation:
Given that:
x(t) = at – bt2+c
a) x(t) = at – bt2+c
Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:
x(t) = 1.4t - 0.06t² + 50
At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m
At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m
The average velocity (v) is given as:
![v=\frac{x(5)-x(0)}{5-0}\\ \\v=\frac{55.5-50}{5-0}=1.1\\ \\v=1.1\ m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%285%29-x%280%29%7D%7B5-0%7D%5C%5C%20%5C%5Cv%3D%5Cfrac%7B55.5-50%7D%7B5-0%7D%3D1.1%5C%5C%20%5C%5Cv%3D1.1%5C%20m%2Fs)
b) x(t) = 1.4t - 0.06t² + 50
At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m
At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m
The average velocity (v) is given as:
![v=\frac{x(10)-x(0)}{10-0}\\ \\v=\frac{58-50}{10-0}=0.8\\ \\v=0.8\ m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%2810%29-x%280%29%7D%7B10-0%7D%5C%5C%20%5C%5Cv%3D%5Cfrac%7B58-50%7D%7B10-0%7D%3D0.8%5C%5C%20%5C%5Cv%3D0.8%5C%20m%2Fs)
c) x(t) = 1.4t - 0.06t² + 50
At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m
At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m
The average velocity (v) is given as:
![v=\frac{x(15)-x(10)}{15-10}\\ \\v=\frac{57.5-58}{15-10}=0.1\\ \\v=0.1\ m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%2815%29-x%2810%29%7D%7B15-10%7D%5C%5C%20%5C%5Cv%3D%5Cfrac%7B57.5-58%7D%7B15-10%7D%3D0.1%5C%5C%20%5C%5Cv%3D0.1%5C%20m%2Fs)