# Finding average speed knowing acceleration

Particle, initially at rest travels in a straight line and its acceleration satisfies $$a=0.1(t-5)^2$$for $0\leq t\leq5$

Find its average speed. during the first$5$ seconds.

I don't expect answers as this clearly is a homework, just a hint what formulas to use to find the answer. Initially I thought that it is $$\frac{\text{final velocity-initial velocity}}{2}$$ but my answer and books answer is different.

HINT

We need to consider an integral average for the speed $|v(t)|$ with $a(t)=\frac{dv(t)}{dt}$ for $t\in[0,5]$, that is

$$\frac{\int_0^5 |v(t)| dt}{5}$$

• but isnt $v=\frac{ds}{dt}$? – Scavenger23 Mar 8 '18 at 20:57
• ops...sorry I fix of course! – gimusi Mar 8 '18 at 20:57
• and that is exactly what I did! I got 4.16666..../5 and the answer in the book is 3.125 – Scavenger23 Mar 8 '18 at 20:59
• @KuderaSebastian did you considered the absolute value? – gimusi Mar 8 '18 at 21:00
• Of course, they did say that the particle starts from the rest, and of course the formula for the acceleration tells you that the velocity is positive in the 5 seconds. – Scavenger23 Mar 8 '18 at 21:20

Your formula is only valid if the acceleration is constant, which it is not in your case. Hint: The average value of a function on $[a,b]$ is \begin{align} \frac{1}{b-a}\int_a^b f(x)\,dx \end{align}

The average speed is the total distance travelled divided by the time.

Integrating once we get an expression for the velocity, namely $$v=\frac{0.1}{3}\left[(t-5)^3+125\right]$$

Integrating again gives displacement, or in this case, distance travelled, since the motion is not reversed in the first $5$ seconds:

$$s=\frac{0.1}{3}\left[\frac 14(t-5)^4+125t\right]_0^5$$

Evaluating this and dividing by $5$ gives the answer $3.125$

Velocity is different from speed. For average speed, you'll need to use the equation of $\frac{d}{\Delta t}$. Where d is the sum of all the absolute displacements (also the total distance travelled). So, the hint here is likely you'll need to consider cases since positive and negative displacement shouldn't cancel out each other as they both contribute to the distance.