If a rock is thrown upward on the planet mars with a velocity of 10 m/s, its height in meters t seconds later is given by y= 10t - 1.86t^2. (a) find the average velocity over the given time intervals: (i) [1, 2] (ii) [1, 1.5] (iii) [1, 1.1] (iv) [1, 1.01] (v) [1, 1.001] (b) estimate the instantaneous velocity when t = 1.

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

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