Respuesta :
Answer:
(a)
[tex]v(t) = \frac{At^2}{2} -\frac{Bt^3}{3}[/tex] --- velocity
[tex]x(t) = \frac{At^3}{6} -\frac{Bt^4}{12} + x(0)[/tex] --- position
(b) The maximum velocity is: 39.0625m/s
Explanation:
Given
[tex]a_x(t) =At -Bt^2[/tex]
[tex]A = 1.5m/s^3[/tex]
[tex]B = 0.120m/s^4[/tex]
Solving (a): The position and velocity as a function
To calculate the change in velocity, we integrate the given acceleration
i.e.
[tex]\int\limits^{v(t)}_{v(0)} {dv} = \int\limits^t_0 {a(t)} \, dt[/tex]
Integrate
[tex]v|\limits^{v(t)}_{v(0)} = \int\limits^t_0 {a(t)} \, dt[/tex]
Substitute v(t) and v(0) for v
[tex]v(t) - v(0) = \int\limits^t_0 {a(t)} \, dt[/tex]
[tex]v(0) = 0[/tex] ---- The initial velocity.
So, we have:
[tex]v(t) - 0 = \int\limits^t_0 {a(t)} \, dt[/tex]
[tex]v(t) = \int\limits^t_0 {a(t)} \, dt[/tex]
Substitute: [tex]a_x(t) =At -Bt^2[/tex]
[tex]v(t) = \int\limits^t_0 {[At -Bt^2]} \, dt[/tex]
Integrate
[tex]v(t) = {\frac{At^2}{2} -\frac{Bt^3}{3}}|\limits^t_0[/tex]
Substitute 0 and t for t
[tex]v(t) = [\frac{At^2}{2} -\frac{Bt^3}{3}] - [\frac{A*0^2}{2} -\frac{B*0^3}{3}][/tex]
[tex]v(t) = [\frac{At^2}{2} -\frac{Bt^3}{3}] - 0[/tex]
[tex]v(t) = \frac{At^2}{2} -\frac{Bt^3}{3}[/tex]
To calculate the change in position, we integrate the calculated velocity
i.e.
[tex]\int\limits^{x(t)}_{x(0)} {dx} = \int\limits^t_0 {v(t)} \, dt[/tex]
Integrate
[tex]x(t)|\limits^{x(t)}_{x(0)} = \int\limits^t_0 {v(t)} \, dt[/tex]
Substitute x(t) and x(0) for x(t)
[tex]x(t) - x(0) = \int\limits^t_0 {v(t)} \, dt[/tex]
Substitute: [tex]v(t) = \frac{At^2}{2} -\frac{Bt^3}{3}[/tex]
[tex]x(t) - x(0) = \int\limits^t_0 {\frac{At^2}{2} -\frac{Bt^3}{3}} \, dt[/tex]
Integrate
[tex]x(t) - x(0) = \frac{At^3}{6} -\frac{Bt^4}{12}}|\limits^t_0[/tex]
Substitute t and 0 for t
[tex]x(t) - x(0) = [\frac{At^3}{6} -\frac{Bt^4}{12}] - [\frac{A*0^3}{6} -\frac{B*0^4}{12}][/tex]
[tex]x(t) - x(0) = [\frac{At^3}{6} -\frac{Bt^4}{12}] - [0][/tex]
[tex]x(t) - x(0) = \frac{At^3}{6} -\frac{Bt^4}{12}[/tex]
Make x(t) the subject
[tex]x(t) = \frac{At^3}{6} -\frac{Bt^4}{12} + x(0)[/tex]
Where x(0) represents the initial position
Solving (b): The maximum velocity
First, we calculate the time at which it attains the maximum height
Set acceleration to 0
[tex]a_x(t) =At -Bt^2[/tex]
[tex]At - Bt^2 = 0[/tex]
Factorize
[tex]t(A - Bt) = 0[/tex]
Split
[tex]t = 0\ or\ A - Bt = 0[/tex]
t can't be 0.
So, we have:
[tex]A - Bt = 0[/tex]
Solve for t
[tex]-Bt = -A[/tex]
[tex]t = \frac{-A}{-B}[/tex]
[tex]t = \frac{A}{B}[/tex]
Substitute: [tex]A = 1.5m/s^3[/tex] and [tex]B = 0.120m/s^4[/tex]
[tex]t = \frac{1.5m/s^3}{0.120m/s^4}[/tex]
[tex]t = \frac{1.5}{0.120}s[/tex]
[tex]t = 12.5s[/tex]
Substitute [tex]t = 12.5s[/tex] in v(t) to get the maximum velocity
[tex]v(t) = \frac{At^2}{2} -\frac{Bt^3}{3}[/tex]
[tex]v(t) = \frac{A*12.5^2}{2} -\frac{B*12.5^3}{3}[/tex]
[tex]v(t) = \frac{156.25A}{2} -\frac{1953.125B}{3}[/tex]
Substitute: [tex]A = 1.5m/s^3[/tex] and [tex]B = 0.120m/s^4[/tex]
[tex]vmax = (156.25*1.5)/2 -(1953.125*0.120)/3[/tex]
[tex]v_{max} = 117.1875 - 78.125[/tex]
[tex]v_{max} = 39.0625[/tex]
