Problem 10 – slowing a car to a full stop
A car is moving at 25m/s when the driver applies the brakes, resulting in constant deceleration. After 5.0 s the speed decreases to 10m/s. What distance does the car travel from the instant the braking begins to the moment the car comes to rest?
Given:
v0 =25m/s – initial speed
v1 = 10m/s – intermediate speed
t1= 5s – time of slowing from v0 to v1
vF = 0m/s final speed (at stop)
We are looking for:
D – distance travel before stopping
To calculate D we must find also:
a – deceleration (acceleration with direction opposite to velocity)
tT – the final time of slowing down.
Analysis. There are a few methods of analyzing and solving the problem. First, just by describing the process of stopping the car.
I) There is an assumption that acceleration (negative) is constant, form the instant braking begins up to a full stop.
The acceleration can be calculated from the first 5 seconds of slowing. Then, knowing the acceleration, the initial velocity and the final velocity (zero at stop), we can calculate the distance traveled from the moment of applying brakes to the moment the car stops.
II) More elegant and general way of solving any problem is:
Count unknown quantity
Write as many equations as there are unknown variables. The equations must be written on the basis of laws governing linear motion with constant acceleration. Then, solve the set of equations and find the unknown. We will follow the II method.
The set of three equations mentioned above is:
D = v0tT – (1/2)atT2 (1)
v1 = v0 – at1 (2)
vF = v0 – atF (3)
From the Eq. 2 and Eq. 3 we get
a = (v0 – v1) / t1 (2a)
a = (v0 – vF) / tT (3a)
As both equations define the same acceleration a, we can write
(v0 – v1) / t1= (v0 – vF) / tT
and from this we can calculate time tT
(4)
Substituting to Eq.1 acceleration a from Eq.2a and time tF from Eq. 4 we get
After a little of elementary algebra you will get
And substituting the numbers given in the text you obtain the distance D = 104.17 m.