Question

A docking shuttle needs to decelerate at a very specific rate, given by a=bt2a=bt2 . if the shuttle is moving at 800m/s800m/s and begins at 30km away, how long will it take to dock?

Answer 1

50 seconds.

Step-by-step explanation:

The given function for acceleration (a) is;

a = bt²

(I) To get the velocity (v) of the shuttle, we integrate the above equation with respect to t as follows;

`\int\limits^x_y {a} \, dt` =
`\int\limits^x_y {bt^(2) } \, dt` --------------------------(ii)

Where;

x and y are the limits of integration.

Integrate both sides of equation (ii)

Δv =
`(bt^(3) )/(3)`

Where;

Δv = velocity at some time t[v(t)] - initial velocity at time t= 0[v(0)]

=> v(t) - v(0) =
`(bt^(3) )/(3)` -------------------(iii)

From the question,

v(0) = initial velocity of the shuttle = 800m/s

Substitute these values into equation (iii) as follows;

v(t) - 800 =
`(bt^(3) )/(3)`

v(t) =
`(bt^(3) )/(3)` + 800 -----------------------(iv)

Also;

(II) To get the distance (s) covered, we integrate equation (iv) as follows;

`\int\limits^x_y {v(t)} \, dt` =
`\int\limits^x_y {((bt^(3) )/(3) + 800) } \, dt` ----------------(v)

Where;

x and y are the limits of integration.

Integrate both sides of equation (v)

Δs =
`(bt^(4) )/(12)` + 800t

Where;

Δs = distance at some time t[s(t)] - initial distance at time t = 0 [v(0)]

=> s(t) - s(0) =
`(bt^(4) )/(12)` + 800t -------------------(vi)

From the question,

s(t) - s(0) = 30km = 30000m

Substitute these values into equation (vi) as follows;

30000 =
`(bt^(4) )/(12)` + 800t

Rewrite the equation as follows;

30000 = (
`(bt^(3) )/(3)` x
`(t )/(4)`) + 800t -----------------------(vii)

(III) Now taking equations(iv) and (vii), let's calculate the time it will take the shuttle to dock as follows;

v(t) =
`(bt^(3) )/(3)` + 800

Where;

v(t) = 0 [since the velocity when the shuttle docks(stops temporarily) is zero]

=> 0 =
`(bt^(3) )/(3)` + 800

=>
`(bt^(3) )/(3)` = -800

Now substitute the value of
`(bt^(3) )/(3)` = -800 into equation (vii) as follows;

30000 = (-800 x
`(t )/(4)`) + 800t

30000 = (
`(-800t)/(4)`) + 800t

Multiply through by 4;

120000 = -800t + 3200t

120000 = 2400t

Solve for t;

t = 120000 / 2400

t = 50

Therefore, the time it will take for the shuttle to dock is 50 seconds.

Answer 2

as it is given here

`a = bt^2`

by integrating both sides

`v_f - v_i = (bt^3)/(3)`

as finally it has to stop

`800 = (bt^3)/(3)`

`2400 = bt^3`

also we know that

`v = 800 - (bt^3)/(3)`

integrating both sides

`x = 800 t - (bt^4)/(12)`

`30000 = 800 t - (2400*t)/(12)`

`30000 = 800t - 200t`

`t = 50 s`

so it will take t = 50 s to cover this distance