Answer:
We know that:
The car stops in 150 m, so this will be the final position.
The acceleration is -7.0m/s^2
I will assume that the initial position is zero.
Then we can write the movement equations as:
Acceleration:
A(t) = -7.0m/s^2
For the velocity, we integrate over time and get:
V(t) = (-7m/s^2)*t + V0.
Now we can integrate again over the time to get the position.
P(t) = (-3.5 m/s^2)*t^2 + V0*t + P0
Where P0 is the initial position,, ut it is equal to zero.
Then the equations that we need to use are:
V(t) = (-7m/s^2)*t + V0.
P(t) = (-3.5 m/s^2)*t^2 + V0*t
The car will reach a complete stop when:
The velocity is equal to zero
The position is equal to 150m
Then we have a system of equations:
0m/s = (-7m/s^2)*t + V0.
150m = (-3.5 m/s^2)*t^2 + V0*t
To solve this, first let's isolate one variable in one of the equations, i wil isolate V0 in the first equation:
V0 = (7m/s^2)*t
Now i can replace this into the second eqation:
150m = (-3.5m/s^2)*t^2 + (7m/s^2)*t^2
150m = (3.5m/s^2)*t^2
t = √( 150/3.5) seconds = 6.54 seconds.
And we know that the initial velocity is:
V0 = (3.5m/s^2)*t
then we can replace the value of t and get:
V0 = (3.5m/s^2)*6.54s = 22.89 m/s
