A car goes from a dead stop to 25m/s in 5s going east. It keeps going then slows down after it reaches 60m/s. It is being drived on level ground. For the first 5s the magnitude of the velocity vector increases at a constant rate. What is the Instantaneous velocity of the car after 3s after first hitting the accelerator from a dead stop?

In my book it says that it is a simple linear function.
It also says that |v|=5t
this is what I don't get...how do they get 5 ?

I know that t=5s |v|=25m/s
but I still don't know how they derive the 5 out ...

since v=q/t or v=d/t
while rearranged the eqzn is q=vt

why I'm confused is that for another eqzn it is explained with acceleration and I think I get it...check my thinking..
if a(t)=10.00m/s2
since a=v/t v=at
|v|(t)= 10.00t

Is the 10.00 from the acceleration like I think or from somewhere else? b/c they have 5 in the other Q but Is it or Isn't it the same eqzn..?

I need help on the concept of where they get the 5 in the problem..

Well, the 5 is the acceleration, but the author of the book is wrong to write the equation like v = 5 t.

"A car goes from a dead stop to 25 m/s in 5 s"

So, the speed increases from 0 to
25 m/s in 5 seconds. The speed at some time t is thus given by:

(25 m/s)/(5 s) t = 5 (m/s^2) t

So, the acceleration is 5 m/s^2. The m/s^2 is the SI unit of the acceleration. You could use the convention to always work in SI units and then omit the units, but that's very confusing, because there is nothing special about SI units. In pratice the quantities you need to calculate with can be given in any units.

If you keep the units in your equations, then you don't have to bother about converting from one set of units to another set of units, the equation will do it for you.

Thanks I was really wondering why there was a 5 but I guess they didn't want to explain how they got it since they didn't explain how to get aceleration yet. They're explanation solution for the problem was "after deriving the eqzn this is what we got" which didn't help at all

Thanks again.