9. The driver of a car going +90.0 km/h suddenly sees the lights of a barrier 40.0 m ahead. It takes the driver 0.75 s to apply the brakes and the average acceleration during braking is -10.0 m/s?.
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a. Does the car hit the barrier?
b. What is the maximum speed at which the car could be moving and not hit the barrier? Assume the acceleration rate does not change.
To determine whether the car hits the barrier, we can calculate the distance it will cover during the reaction time and compare it with the distance to the barrier.
Distance covered during reaction time:
d_reaction = v_initial * t_reaction
= (90.0 km/h) * (0.75 s)
= (90.0 * 1000 m / 3600 s) * (0.75 s)
= 25 m
The distance to the barrier is given as 40.0 m.
Since the distance covered during the reaction time (25 m) is less than the distance to the barrier (40.0 m), the car will hit the barrier.
To find the maximum speed at which the car could be moving and not hit the barrier, we can use the equation of motion:
d = v_initial * t + (1/2) * a * t^2
Substituting the given values:
40.0 m = v_initial * (0.75 s) + (1/2) * (-10.0 m/s^2) * (0.75 s)^2
Rearranging the equation:
(1/2) * (-10.0 m/s^2) * (0.75 s)^2 + v_initial * (0.75 s) - 40.0 m = 0
Solving this quadratic equation using the quadratic formula:
v_initial = (-0.75 s ± √((-0.75 s)^2 - 4 * (1/2) * (-10.0 m/s^2) * (-40.0 m))) / (2 * (1/2) * (-10.0 m/s^2))
v_initial = (-0.75 s ± √(0.5625 s^2 + 800 m^2/s^2)) / (-10.0 m/s^2)
v_initial = (-0.75 s ± √(800.5625 m^2/s^2)) / (-10.0 m/s^2)
Since speed cannot be negative, we consider the positive solution:
v_initial = (-0.75 s + √(800.5625 m^2/s^2)) / (-10.0 m/s^2)
v_initial = (-0.75 s + 28.3 m/s) / (-10.0 m/s^2)
v_initial = (27.55 m/s) / (10.0 m/s^2)
v_initial ≈ 2.76 m/s
Therefore, the maximum speed at which the car could be moving and not hit the barrier is approximately 2.76 m/s or 9.94 km/h.
a. To determine if the car hits the barrier, we need to calculate the distance the car will travel during the 0.75 s it takes to apply the brakes.
We can use the equation:
d = vi*t + (1/2)*a*t^2
where d is the distance traveled, vi is the initial velocity, t is the time, and a is the acceleration.
In this case, vi is +90.0 km/h (since the car is moving forward), t is 0.75 s, and a is -10.0 m/s^2 (since the acceleration is in the opposite direction of motion).
First, we need to convert the initial velocity from km/h to m/s:
+90.0 km/h = (90.0 km/h) * (1000 m/km) * (1/3600 h/s) = 25.0 m/s
Now, we can plug the values into the equation:
d = (25.0 m/s)(0.75 s) + (1/2)(-10.0 m/s^2)(0.75 s)^2
= (18.75 m) - (2.8125 m)
= 15.9375 m
Therefore, the car will travel a distance of 15.9375 m during the 0.75 s it takes to apply the brakes. Since this distance is less than the distance to the barrier (40.0 m), the car does not hit the barrier.
b. To find the maximum speed at which the car could be moving and not hit the barrier, we need to determine the distance the car can stop within.
Using the equation:
vf^2 = vi^2 + 2*a*d
where vf is the final velocity (0 m/s), vi is the initial velocity, a is the acceleration, and d is the distance.
Rearranging the equation to solve for vi, we have:
vi = sqrt(vf^2 - 2*a*d)
Plugging in the values:
vf = 0 m/s, a = -10.0 m/s^2, and d = 40.0 m
vi = sqrt((0 m/s)^2 - 2*(-10.0 m/s^2)(40.0 m))
= sqrt(0 - (-800.0) m^2/s^2)
= sqrt(800.0) m/s
≈ 28.3 m/s
Therefore, the maximum speed at which the car could be moving and not hit the barrier is approximately 28.3 m/s.