Answer:
Suppose that the acceleration is a constant, a.
a(t) = a.
To write the velocity equation, we must integrate over time, and the constant of integration will be equal to the initial velocity, in this case is 110km/h.
v(t) = a*t + 110km/h
And we know that at t = 5s, the car was brought to stop, so the velocity must be zero.
v(5s) = 0 = a*5s + 110km/h.
a = (110km/h)*(1/5s)
now we have that:
1 hour = 3600 seconds.
1km = 1000m
then:
110km/h = (110*1000/3600)m/s = 30.56 m/s
Then we have:
a = (-30.55 m/s)/5s = -6.11 m/s^2
Now the velocity equation is:
v(t) = -6.11m/s^2*t + 30.56m/s
To write the positon equation we must integrate over time again, we can get:
p(t) = (1/2)*(-6.11m/s^2)*t^2 + (30.56m/s)*t + p0
Where p0 is the initial position, here i will assume that is zero, because it does no really mater.
The total displacement of the car will be equal to p(5s)
p(5s) = (1/2)*(-6.11m/s^2)*(5s)^2 + (30.56m/s)*(5s) = 76.425 meters.
