Time to warm up!

First, let's write everything in terms of meters and seconds.

72 kilometers per hour is equal to 72,000 meters per hour, and 20 meters per second. Therefore, the average velocity of the car is 20 meters per second.

We can model v(t) as a piecewise function, with v in meters/second and t in seconds.

u represents the time the car begins its uniform (of constant velocity) motion. g represents the time the car ends its uniform motion.

$\displaystyle v(t)= 5t, \: \: 0 \leq t \leq u$

$\: \: \: \: v(u), \: \: u < t \leq g$

$\: \: \: \: -5(t-25), \: \: g < t \leq 25$

The uniform portion of the car's motion must have a constant velocity, so it will simply continue at the velocity reached at the end of the interval where the car is accelerating, which would be the velocity at the later endpoint, or v(u).

We know that the car's velocity is 0 when t=25, so the third portion of our piecewise function for v(t) is =-5(t-25), since we can perform a simple horizontal shift to the model derived from $\Delta v = a \times \Delta t$ , since (25,0) is the horizontal intercept of that segment.

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IF you can easily see that the time accelerating must be the same as the time "decelerating", since the magnitudes are equal and the net change in velocity is 0, feel free to skip ahead quite a bit.

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We also know that since the car's final velocity (when t=25), is the same as its initial velocity (when t=0), the net change in its velocity from t [0,25] is 0 m/s.

Since the acceleration from t[0,u] is 5, and the acceleration from t[g,25] is -5, we can use our equation for change in velocity: $\Delta v = a \times \Delta t$

For t[0,u]: $\Delta v = 5 \times u$

For t[g,25]: $\Delta v= -5 \times (25-g)$

For the rest of the interval t[0,25], the acceleration is 0, so the net change in velocity over that interval is composed of the sum of the two above $\Delta V$ values.

For t[0,25]: $\Delta v = 5 \times u+ =5 \times (25-g) =0$ , as determined above.

Therefore, $u=25-g$ and $g=25-u$

Now our v(t) function can be simplified quite a bit:

$\displaystyle v(t)= 5t, \: \: 0 \leq t \leq u$

$\displaystyle \: \: \: \: v(u), \: \: u < t \leq (25-u)$

$\displaystyle \: \: \: \: -5(t-25), \: \: (25-u) < t \leq 25$

The average velocity is 20 meters per second, and one way to define the average velocity is by manipulating a definite integral. Like the familiar arithmetic mean, this sums the values, and divides by the "number" of values being averaged.

$\displaystyle \frac{1}{25} \times \int_0^{25}v(t)dt=20$

$\displaystyle \int_0^{25}v(t)dt=500=500$

$\displaystyle \int_0^uv(t)dt+\int_u^{25-u}v(t)dt+\int_{25-u}^{25}v(t)dt=500$

Referring to our piecewise v(t) function: $\displaystyle \int_0^u[5t]dt+\int_u^{25-u}[v(u)]dt+\int_{25-u}^{25}[-5(t-25)]dt=500$

$\displaystyle v(u) = 5 \times u$

Evaluating the integrals in terms of u: $\displaystyle \frac{5}{2} \times t^2 |_0^u +t \times 5u|_u^{25-u} + -\frac{5}{2} t(t-50)|_{25-u}^{25}=500$

$\displaystyle \frac{5}{2} \times u^2 +(25-u) \times 5u -u \times 5u + [-\frac{5}{2} 25(25-50)- -\frac{5}{2} (25-u)(25-u-50)]=500$

Through a large amount of algebra (yes, we definitely could've avoided most of this by rewriting the original integral as the area of a trapezoid $\displaystyle \int_0^{25}v(t)dt=500=A=\frac{b_1+b_2}{2}\times h=\frac{(g-u)+25}{2} \times v(u)=\frac{(25-u-u)+25}{2} \times 5u$ ) , you eventually get the solutions u=5 seconds or u=20 seconds. The solution that makes sense here is u=5 seconds, since if the car began its uniform motion at t=20 seconds, then the car would've been accelerating for 20 seconds, leaving only 5 seconds for uniform motion & "deceleration".

To find the length of the interval of uniform motion, we simply find the difference between the starting time of uniform motion and its end time.

g=25-u

u=5

g-u=25-5-5= 15 seconds

Here is a graph of the car's velocity against time: