What is the average speed, on t in [0,5], of an object that is moving at 7 m/s at t=0 and accelerates at a rate of a(t) =2t^2+2 on t in [0,2]?

1 Answer
Jun 10, 2017

32.8"m"/"s"

Explanation:

The average speed of the object is represented by the equation

overbrace(v_"av-x")^"speed" = "total distance traveled"/"time interval"

Since the acceleration is always positive, the total distance traveled is equal to the total displacement, so the average speed is the same as the average velocity, so we can use the equation

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

We must therefore find the total displacement of the object after t = 5 "s".

To find the position equation from the acceleration equation, we need to do an integration. We need to integrate the acceleration equation twice because position is the second integral of acceleration (velocity lies in between, as the first integral). We can use the equation

v_x = v_(0x) + int_0^t a_xdt

to find the velocity equation, and then integrate the velocity equation to find the position equation. Knowing that

int t^n = 1/(n+1)t^(n+1)

the integral of color(red)(2)t^(color(blue)(2) is

(1/(color(blue)(2)+1))(color(red)(2))t^(color(blue)(2)+1) = 2/3t^3

and the integral of color(red)(2, which is 2t^(color(blue)(0), is

(1/(color(blue)(0)+1))(color(red)(2))t^(color(blue)(0)+1) = 2t

The initial velocity is 7"m"/"s". Plugging these values into the integral equation, we have

v_x = 7 + 2/3t^3 + 2t

(excluding units)

The equation for position of time from velocity is

x = x_0 + int_0^t v_xdt

Its starting position is assumed to be x = 0. Using the same technique as before, we have

x = 0 + 7t + 1/6t^4 + t^2

(excluding units)

Now, all we have to do is find the position at t = 5 "s". Plugging in 5 for t to this equation, we have

x = 0 + 7(5) + 1/6(5)^4 + (5)^2 = color(purple)(164.2 color(purple)("m"

Finally, let's plug this back into our average velocity equation:

v_"av-x" = (color(purple)(164.2"m"))/(5"s") = color(green)(32.8"m"/"s"