A car travelling at 100 kmh-1 decelerates uniformly at 1.8 ms-2.
Calculate:

a.the distance it goes before it stops;


b.the time it takes to stop; and


c.the distance it travels during the first and third seconds.

Hello pj444!
These are the answerst to your questions, I will first answer question
b, as its solution makes it easy to solve question a.

b. First of all, to have everything in comparable units, we must find
out what is 100kmh-1 in terms of ms-1 (meters per second). This is
fairly easy:

100km/h = 100.000 m/h = (100.00/3600) m/s = 27.7778 m/s

Now, a deceleration of 1.8 ms-2 means that velocity is decreasing at a
rate of 1.8 m/s per second. Thus if we define V=velocity and
s=seconds, we can know what the velocity of the car is after s seconds
from the initial time with the following formula:

Velocity(s) = ( 27.7778 - 1.8*s ) m/s

Now, we can find out how long the car takes to stop. That is, we must
find how many seconds must pass so that Velocity reaches 0. Using the
above formula,

0 = 27.7778 - 1.8*s
1.8*s = 27.7778
s = 15.4321

That's the answer to question b: the car takes 15.4321 seconds to stop
completely.


a. To find the distance the car has gone in the 15.4321 seconds, and
since this is a case of uniformly decreasing velocity, we can use the
averge velocity of the car:

Initial velocity (V0) = 27.7778 m/s
Final velocity (Vf) = 0 m/s
Average velocity = (V0+Vf) / 2 = 13.8889 m/s

So 13.8889 m/s is the average velocity of the car in the 15.4321
seconds that pass before it stops. Now we may use the formula

distance = velocity * time

distance = 13.8889 m/s * 15.4321 s = 214.3349 m

That answers part a.


c. The answer to this question can be found in the same way question a
is answered.

* First Second
The velocity at the beggining of the first second is 27.7778 m/s.
The velocity at the end of the first second is 
27.7778 - 1.8 = 25.9778 m/s

Average velocity is then 
(27.7778 + 25.9778) / 2 = 26.8778 m/s

So the distance is 26.8778 m/s * 1 s = 26.8778 m

The same procedure can be used for the third second, obtaining that
the distance the car travels during the 3rd second is 21.4778 m/s.


You may want to ckeck the following page for some useful formulas
regarding Constant Acceleration in 1 dimension
http://mcasco.com/p1consta.html

Finally, note that exercises a and c can also be solved using calculus
(specifically, integration), you will arrive to the same results. To
find the distance the car travels before it stops, you would have to
integrate Velocity(s) in the interval [0, 15.4321]. In exercise c, you
would integrate the same function in the intervals [0, 1] and [3, 4].
The results are exactly the same.

Hope this helps!


Best regards,
elmarto