Problem 6: Variable Velocity
We are given the velocity function \(v(t)=t^2+2t-5\) and the initial position \(x(0)=4\).
We want:
- The position at \(t=3\), i.e. \(x(3)\).
- The acceleration at \(t=3\), i.e. \(a(3)\).
1) Position from velocity theory
Velocity is the derivative of position:
\[
v(t)=\frac{dx}{dt}
\]
So:
\[
\frac{dx}{dt}=t^2+2t-5
\]
Integrate both sides with respect to \(t\):
\[
x(t)=\int (t^2+2t-5)\,dt
\]
Compute the integral term-by-term:
\[
\int t^2\,dt=\frac{t^3}{3},\qquad
\int 2t\,dt=t^2,\qquad
\int (-5)\,dt=-5t
\]
So the general position function is:
\[
x(t)=\frac{t^3}{3}+t^2-5t+C
\]
Use the initial condition \(x(0)=4\):
\[
x(0)=\frac{0^3}{3}+0^2-5\cdot 0 + C = C = 4
\]
Therefore:
\[
x(t)=\frac{t^3}{3}+t^2-5t+4
\]
Now evaluate at \(t=3\):
\[
x(3)=\frac{3^3}{3}+3^2-5\cdot 3+4
=\frac{27}{3}+9-15+4
=9+9-15+4
\]
Compute: $$ 9+9=18,\qquad 18-15=3,\qquad 3+4=7 $$
So:
\[
x(3)=7
\]
2) Acceleration from velocity
Acceleration is the derivative of velocity:
\[
a(t)=\frac{dv}{dt}
\]
Differentiate:
\[
v(t)=t^2+2t-5
\quad\Rightarrow\quad
a(t)=2t+2
\]
Evaluate at \(t=3\):
\[
a(3)=2\cdot 3 + 2 = 8
\]
Final answers
\[
x(3)=7
\]
\[
a(3)=8
\]
Plot the velocity and acceleration
