Question

What is the distance that a car travels if it was brought to stop in 5 seconds and if it was traveling at 110 Km/h

Answer 1

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.