Question

What is the average speed of an object that is still at `t=0` and accelerates at a rate of `a(t) = 5t+1` from `t in [0, 6]`?

Answer 1

`33"m"/"s"`

Explanation:

To find the average speed of the object, we have to know both the total distance traveled (`Deltax`) and the time (`6"s"`, according to the given interval).

We need to integrate this acceleration equation in order to find the position equation. The integral of `t^n` (`n` being `1` in this case, the exponent of `t` in the "`5t`" quantity) is denoted `int t^ndt`, and is equal to `1/(n+1)t^(n+1)`, which in this case is

`1/(1+1)t^(1+1) = 1/2t^2`

Doing the same for the second quantity (`1`), we have

`1/(0+1)t^(0+1) = 1t^1 = t`

Therefore, the first integral of the acceleration equation, which is the velocity equation, is

`v_x(t) = v_(0x) + 1/2(5)t^2 + t = 5/2t^2 + t`

because the initial velocity and position are assumed to be `0`.

Since this is the velocity equation, we need to integrate this again to find the position equation. Following the same procedure as before, our result is

`x(t) = x_0 + 1/3(5/2)t^3+1/2t^2 = 5/6t^3 + 1/2t^2`

Now that we have our position equation, let's find the position of the particle at the time `t = 6"s"`:

`x(6) = 5/6(6)^3+ 1/2(6)^2 = color(red)(198"m"`

Since the velocity and position are always increasing here, the average speed is the same as the average velocity, which is given by the equation

`v_(av-x) = (Deltax)/(Deltat)`

Since our initial position and time are both `0`, the average speed of the object over the interval `t in [0,6]` is

`v_(av-x) = (color(red)(198"m"))/(6"s") = color(blue)(33"m"/"s"`