Answer:
a)
i) v = 4.42 m/s
ii) v = 5.36 m/s
iii) v = 6.1 m/s
iv) v = 6.26 m/s
v) v = 6.28 m/s
b) The instantaneous velocity at t = 1 is 6.28 m/s
Explanation:
a) The average velocity is the variation of the position over time. It is expressed as follows:
v = Δy/Δt
Where
v = average velocity
Δy = displacement = final position - initial position
Δt = variation of time = final time - initial time
i) Let´s find the position at both times and then apply the equation for the average velocity:
y(t) = 10 · t - 1.86 · t²
y(1 s) = 10 m/s · 1 s - 1.86 m/s² · (1 s)²
y = 8.14 m
y (2 s) = 10 m/s · 2 s - 1.86 m/s² · (2 s)²
y = 12.56 m
Then, the average velocity will be:
v = final position - initial position / final time - initial time
v = 12.56 m - 8.14 m / 2 s - 1 s = 4.42 m/s
ii) We proceed in the same way as in i)
y(1.5 s) = 10 m/s · 1.5 s - 1.86 m/s² · (1.5 s)²
y = 10.82 m
v = 10.82 m - 8.14 m / 1.5 s - 1 s = 5.36 m/s
iii)
y(1.1 s) = 10 m/s · 1.1 s - 1.86 m/s² · (1.1 s)²
y = 8.75 m
v = 8.75 m - 8.14 m / 1.1 s - 1 s = 6.1 m/s
iv)
y(1.01 s) = 10 m/s · 1.01 s - 1.86 m/s² · (1.01 s)²
y = 8.20 m
v = 8.20 m - 8.14 m / 1.01 s - 1 s = 6 m/s ( 6.26 m/s without rounding the y-final value)
v)
y(1.001 s) = 10 m/s · 1.001 s - 1.86 m/s² · (1.001 s)²
y = 8.146
v = 8.146 m - 8.14 m / 1.001 s - 1 s = 6 m/s (6.28 m/s without rounding the value of y-final)
b) The instantaneous velocity is given by the derivative of the position function:
y = 10 · t - 1.86 · t²
dy/dt = 10 - 2 · 1.86 · t = 10 - 3.72 · t
At t = 1
v = 10 m/s - 3.72 m/s² · 1 s = 6.28 m/s
