Initial value problem (solve for velocity)

Question

Gravity is a constant value of -10 metres per second squared. An objects initial velocity is 10 metres per second. Determine an expression for velocity.

So...

g = 10 m/s^2

g = v'(t)

v(0) = 10

and I need to solve an initial value problem, to determine v(t)

Would really appreciate some help on how to start this.

Answer 1

The velocity decreases by 10 meters per second every second; this is what it means for the acceleration to be -10m/s^2.

So $v(0) = 10$, and to determine $v(t)$, just consider how much time has elapsed and how the acceleration would affect velocity during that time.

Answer 2

You have already set up the problem.

In other notation, you basically have that $$\frac{dv}{dt} = g$$ with $v(0)= 10$.

This is a differential equation. Note $g$ is just a constant.

Can you solve differential equations? This one is especially simple.

Hint: try integrating both sides with respect to $t$ and don't forget your constant of integration. That constant can be found using your initial condition to fully solve the IVP.

Answer 3

Since we know that $a(t) = v'(t)$, then integrating both sides with respect to $t$, time, gives us an equation for $v(t)$:

$$v(t) = \int a(t) \, \mathrm dt$$

Since, in this case, $a$ is taken to be a constant, we get

$$v(t) = \int a \, \mathrm dt = at+C = at+v_0$$

where the constant of integration represents our initial velocity, $v_0$. Here, $a = -10$ and $v_0 = 10$ (ignoring the units), so you simply plug those in.

Also, keep in mind that the only reason switching $C$ with $v_0$ is justified is that we're dealing with polynomials. Here, $v_0 = v(0) = a(0)+C = C$, so the switch is justified. Otherwise, you have to find $C$ by using $v_0 = v(0)$, where $v_0$ is the initial velocity given by the question and $v(0)$ is your function evaluated at $t = 0$.

As an extra note, this is one of the basic kinematic formulas, where a constant value for the function $a(t)$ will yield $v(t) = at+v_0$ and $x(t) = \frac{1}{2}at^2+v_0t+x_0$ for functions of velocity and position, respectively. (As an extra exercise, maybe you can try using the exact same method here for showing how the function for position is obtained.)