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:
$\ \ \ \ a(t) = t + 4;\ \  v(0) = 5,\ 0 &lt= t &lt= 10$
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.
 A: 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
$$
s=\int_{0}^{10} v(t)\,dt.
$$
A: 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...
