What is the maximum speed at which a car could be moving and not hit a barrier 35.0 m ahead if the average acceleration during braking is -10.0 m/s2 and it takes the driver 0.90 s before he applies the brakes?
subtract from the 35 m the distance during reactiontime (v*.9) so the stopping distance is 35-.9v.
That is equal to d in ...
vf^2=v^2+2ad
vf is zero, solve for v.
Do I use the acceleration for the velocity or do I figure out the velocity from the informatin given?
To answer the question, you need to use the information given and figure out the velocity. Here's how you can proceed:
First, calculate the distance during the reaction time: (velocity * reaction time) = (v * 0.9).
Next, subtract the distance during reaction time from the total distance to get the stopping distance: 35.0 m - (v * 0.9).
Using the equation vf^2 = v^2 + 2ad, where vf is the final velocity (which is zero because the car comes to a stop), v is the initial velocity, a is the acceleration, and d is the stopping distance, you can substitute the values:
0 = v^2 + 2(-10.0 m/s^2) * (35.0 m - (v * 0.9)).
Simplify the equation:
0 = v^2 - 20.0 m/s^2 * (35.0 m - 0.9v).
Expand:
0 = v^2 - 700.0 m/s^2 + 18.0 m/s^2 * v.
Rearrange the equation:
0 = v^2 + 18.0 m/s^2 * v - 700.0 m/s^2.
Now, solve this quadratic equation for v.
You can use the quadratic formula: v = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = 1, b = 18.0 m/s^2, and c = -700.0 m/s^2.
Substitute these values into the quadratic formula and solve for v.
Note that there will be two solutions, but you only need the positive value because the speed cannot be negative in this context.
Once you find the positive value of v, that will be the maximum speed at which the car could be moving without hitting the barrier 35.0 m ahead.