# Finding time traveled from acceleration function

I don't know how to do this, it doesn't make sense to me and there are no examples in the book.

The acceleration function and the initial velocity are given for a particle moving along a line:

$\ \ \ \ a(t) = t + 4;\ \ v(0) = 5,\ 0 <= t <= 10$

Find:

$\ \ \$a) the velocity at time $t$.

and

$\ \ \$b) the distance traveled during the given time interval.

I am not sure how $t+ 4$ could equal $5$ at $0$. It doesn't make sense.

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The acceleration function is $$\tag{1}a(t)=4+t;\ \ 0\le t\le10.$$ So, for example, the acceleration at $t=0$ is $a(0)=1+0=1$.

To find the velocity function, recall that the derivative of the velocity function is the acceleration function. So to find the velocity function, we can first find an antiderivative of the acceleration function $(1)$. An antiderivative of $a(t)=4+t$ is $4t+{1\over2}t^2+C$. So, for $0\le t\le10$ $$\tag{2} v(t)=4t+\textstyle{1\over2}t^2+C$$ for some constant $C$. We need to find the value of $C$. To do this, we use the information that $v(0)=5$. Using equation ${2}$, we have $$5=v(0)=4\cdot0 +\textstyle{1\over 2} \cdot0^2+C\ \ \Rightarrow\ \ C=5.$$ Thus, the velocity function is $$\tag{3}v(t)=4t+\textstyle{1\over2}t^2+5;\ \ 0\le t\le10.$$

Alternatively, you could just compute the velocity function via, for $0\le t\le 10$ $$v(t)-v(0) =\int_0^t 4+s\,ds= (4s+\textstyle{1\over2}s^2)|_0^t=4t+\textstyle{1\over2}t^2,$$ which gives $(3)$ by moving $-v(0)=-5$ in the equation above to the other side.

Towards finding the distance traveled, we could first find the displacement function. The derivative of the displacement function is the velocity function. So the displacement function, $d$, can be obtained by finding an antiderivative of $(3)$: $$d(t)=2t^2+\textstyle{1\over 6}t^3+5t+D, \ \ 0\le t\le10,$$ for some constant $D$. We can't find the value of $D$ here, but it's not needed to find the distance traveled from $t=0$ to $t=10$. Since $v(t)$ is nonnegative for $0\le t\le10$, the object is always moving to the right; so the total distance traveled from $t=0$ to $t=10$ is $$d(10)-d(0) =[ 200+(1000/6) +50+D] -[ 0+0+0-D] = 200+(1000/6) +50.$$

Alternatively, to find the distance traveled, you could compute $\int_0^{10} v(t)\,dt$; which would lead to exactly the same expression as above.

Here it is important to realize that $v(t)\ge0$ for $0\le t\le 10$. For a general velocity function $v$, the expression $\int_a^b v(t)\, dt$ gives the displacement from $t=a$ to $t=b$. The total distance traveled is $\int_a^b |v(t)|\, dt$. The two can of course be different...

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This is just a hint: $t+4$ shouldn't be equal to 5 since the former is acceleration and the latter is the speed. The connection is, as you certainly know, $v'(t)=a(t)$.

Update:

We know that the acceleration is the derivative of the velocity function: $$v'(t)=a(t).$$ From this formula one has $$v(t)=v(t_0)+\int_{t_0}^t a(\tau)\,d\tau,$$ or, for this particular problem $$v(t)=5+t^2/2+4t.$$ Which gives for $t=10$ $$v(10)=5+100/2+40=95.$$ The distance can be found as $$s=\int_{0}^{10} v(t)\,dt.$$

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Is that from the fundamental theorem of calculus? I don't quite understand it, we didn't really go over it in class we just memorized the formulas. – user138246 Apr 22 '12 at 18:13
Yes. The speed $v(t)$ is the derivative of the distance $v(t)=s'(t)$. And the acceleration is the derivative of the speed $a(t)=v'(t)=s''(t)$. Therefore, the distance is the integral of speed and the velocity is the integral of acceleration. To find velocity, e.g., you need to evaluate the integral $v(t)=\int_{t_0}^t a(\tau)\,d\tau+v(t_0)=\int_{0}^t(\tau+4)\,d\tau+5$. – Artem Apr 22 '12 at 18:17
How do you I know which formula is speed, distance, acceleration or whatever? – user138246 Apr 22 '12 at 18:20
The notation: $a(t)$ is the acceleration; $v(t)$ is the velocity (speed); $s(t)$ is the distance. – Artem Apr 22 '12 at 18:21
So I know that t = 1, what does that tell me? For $t^2 + 4t = 5 t =1$ – user138246 Apr 22 '12 at 18:23