Respuesta :
as it is given here
[tex]a = bt^2[/tex]
by integrating both sides
[tex]v_f - v_i = \frac{bt^3}{3}[/tex]
as finally it has to stop
[tex]800 = \frac{bt^3}{3}[/tex]
[tex]2400 = bt^3[/tex]
also we know that
[tex]v = 800 - \frac{bt^3}{3}[/tex]
integrating both sides
[tex]x = 800 t - \frac{bt^4}{12}[/tex]
[tex]30000 = 800 t - \frac{2400*t}{12}[/tex]
[tex]30000 = 800t - 200t[/tex]
[tex]t = 50 s[/tex]
so it will take t = 50 s to cover this distance
Answer:
50 seconds.
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;
[tex]\int\limits^x_y {a} \, dt[/tex] = [tex]\int\limits^x_y {bt^{2} } \, dt[/tex] --------------------------(ii)
Where;
x and y are the limits of integration.
Integrate both sides of equation (ii)
Δv = [tex]\frac{bt^{3} }{3}[/tex]
Where;
Δv = velocity at some time t[v(t)] - initial velocity at time t= 0[v(0)]
=> v(t) - v(0) = [tex]\frac{bt^{3} }{3}[/tex] -------------------(iii)
From the question,
v(0) = initial velocity of the shuttle = 800m/s
Substitute these values into equation (iii) as follows;
v(t) - 800 = [tex]\frac{bt^{3} }{3}[/tex]
v(t) = [tex]\frac{bt^{3} }{3}[/tex] + 800 -----------------------(iv)
Also;
(II) To get the distance (s) covered, we integrate equation (iv) as follows;
[tex]\int\limits^x_y {v(t)} \, dt[/tex] = [tex]\int\limits^x_y {(\frac{bt^{3} }{3} + 800) } \, dt[/tex] ----------------(v)
Where;
x and y are the limits of integration.
Integrate both sides of equation (v)
Δs = [tex]\frac{bt^{4} }{12}[/tex] + 800t
Where;
Δs = distance at some time t[s(t)] - initial distance at time t = 0 [v(0)]
=> s(t) - s(0) = [tex]\frac{bt^{4} }{12}[/tex] + 800t -------------------(vi)
From the question,
s(t) - s(0) = 30km = 30000m
Substitute these values into equation (vi) as follows;
30000 = [tex]\frac{bt^{4} }{12}[/tex] + 800t
Rewrite the equation as follows;
30000 = ([tex]\frac{bt^{3} }{3}[/tex] x [tex]\frac{t }{4}[/tex]) + 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) = [tex]\frac{bt^{3} }{3}[/tex] + 800
Where;
v(t) = 0 [since the velocity when the shuttle docks(stops temporarily) is zero]
=> 0 = [tex]\frac{bt^{3} }{3}[/tex] + 800
=> [tex]\frac{bt^{3} }{3}[/tex] = -800
Now substitute the value of [tex]\frac{bt^{3} }{3}[/tex] = -800 into equation (vii) as follows;
30000 = (-800 x [tex]\frac{t }{4}[/tex]) + 800t
30000 = ([tex]\frac{-800t}{4}[/tex]) + 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.
