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.
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?.
+
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.
1 answer
1 answer