1.) A rail gun uses electromagnetic energy to accelerate objects quickly over a short distance. In an experiment, a 2.00 kg projectile remains on the rails of the gun for only 2.10x10-2 s, but in that time it goes from rest to a velocity of 4.00×103 m/s. What is the average acceleration of the projectile?
2.) The city is trying to figure out how long the traffic light should stay yellow at an intersection. The speed limit on the road is 45.0 km/h and the intersection is 23.0 m wide. A car is traveling at the speed limit in the positive direction and can brake with an acceleration of -5.20 m/s2. (a) If the car is to stop on the white line, before entering the intersection, what is the minimum distance from the line at which the driver must apply the brakes? (b) How long should the traffic light stay yellow so that if the car is just closer than that minimum distance when the light turns yellow, it can safely cross the intersection without having to speed up?
a. acceleration=changeinVelocity/time
b. a. vf^2=vi^2+2ad solve for d.
b. time=(d+23 )/velocity
change velocity to m/s first.
1) To find the average acceleration of the projectile, we can use the formula:
average acceleration = (final velocity - initial velocity) / time
Given:
mass of the projectile (m) = 2.00 kg
time (t) = 2.10x10^-2 s
final velocity (v_f) = 4.00x10^3 m/s
initial velocity (v_i) = 0 (since it starts from rest)
Plugging in the values, we get:
average acceleration = (4.00x10^3 m/s - 0) / (2.10x10^-2 s)
= 1.90x10^5 m/s^2
Therefore, the average acceleration of the projectile is 1.90x10^5 m/s^2.
2) (a) To find the minimum distance from the white line at which the driver must apply the brakes, we can use the equation of motion:
final velocity squared = initial velocity squared + 2 * acceleration * distance
Given:
speed limit (v_i) = 45.0 km/h (which we convert to m/s by multiplying by 1000/3600)
acceleration (a) = -5.20 m/s^2
distance (d) = ?
Since the car is coming to a stop, the final velocity (v_f) is 0.
Plugging in the values, we get:
0 = (45.0 m/s)^2 + 2 * (-5.20 m/s^2) * d
Simplifying the equation, we find:
45.0^2 = 10.4 * d
2025 = 10.4 * d
d = 2025 / 10.4
Solving the division, we get:
d ≈ 194.71 m
Therefore, the minimum distance from the white line at which the driver must apply the brakes is approximately 194.71 m.
(b) To find the duration of the yellow light, we can use the formula of motion again:
d = initial velocity * t + (1/2) * acceleration * t^2
Given:
speed limit (v_i) = 45.0 km/h (which we convert to m/s by multiplying by 1000/3600)
acceleration (a) = -5.20 m/s^2
distance (d) = 194.71 m
time (t) = ?
Adding the distance (d) and the initial velocity (v_i) to the equation, we get:
d + v_i * t = (1/2) * a * t^2
194.71 m + (45.0 m/s) * t = (1/2) * (-5.20 m/s^2) * t^2
Rearranging the equation, we find:
0 = (1/2) * (-5.20 m/s^2) * t^2 + (45.0 m/s) * t - 194.71 m
This is a quadratic equation, and we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
t = [-(45.0 m/s) ± √((45.0 m/s)^2 - 4 * (1/2) * (-5.20 m/s^2) * (-194.71 m))] / [2 * (1/2) * (-5.20 m/s^2)]
Simplifying the equation, we find:
t = [-(45.0 m/s) ± √(2025 m^2/s^2 + 5097.2424 m^2/s^2)] / (-5.20 m/s^2)
Solving the equation, we get two values for t:
t ≈ 7.77 s or t ≈ -8.13 s
Since time cannot be negative, we discard the negative value. Therefore, the duration of the yellow light should be approximately 7.77 seconds.
To summarize:
(a) The minimum distance from the white line at which the driver must apply the brakes is approximately 194.71 meters.
(b) The traffic light should stay yellow for approximately 7.77 seconds so that the car can safely cross the intersection without having to speed up.