Speedy Sue, driving at 35.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 175 m ahead traveling at 5.50 m/s. Sue applies her brakes but can accelerate only at −1.40 m/s2 because the road is wet. Will there be a collision?
Yes
No
If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue's car and the van. (If no, enter "0" for the time.)
distance - ???
time - ????
using Calculus:
a = -1.4 m/s^2
v = -1.4t + c
when t = 0, v = 35, so c = 35
v = -1.4t + 35
when she stops (with nothing in front of her), v = 0
t = 35/1.4 = 25 seconds
so it would take her 25 seconds to stop
s = -.7t^2 + 35t + k
when t = 0, s = 0, so k = 0
s = -.7t^2 + 35t
so in the 25 seconds needed to stop, she would have gone
-.7(25^2) + 35(25) or 437.5 m
in the meantime, the van would have covered an additional 25(5.5)
or 137.5 m, plus the original distance of 175 ahead would put the van 312.5 m ahead of the place when she first applied the brakes.
So she will rear end the van!
when does this happen?
when -.7t^2 + 35t = 175+5.5t
.7t^2 - 29.5t + 175 = 0
t = 50/7 seconds or 35 seconds, rejecting the 35 s,
clearly she will rear end at 50/7 s or appr 7.14 seconds
distance travelled by both = 175 + 5.5(50/7) = appr 214.3 m
To determine if there will be a collision, we need to find out if Sue's car can come to a complete stop before reaching the van.
First, let's calculate the time it will take for Sue's car to reach the van:
Using the equation for the distance traveled with constant acceleration:
distance = initial velocity * time + 0.5 * acceleration * time^2
For Sue's car:
initial velocity = 35.0 m/s
acceleration = -1.40 m/s^2
distance = 175 m
175 = 35.0 * time + 0.5 * (-1.40) * time^2
Simplifying the equation, we get:
-0.7 * time^2 + 35.0 * time - 175 = 0
To solve this quadratic equation, we can use the quadratic formula:
time = (-b ± sqrt(b^2 - 4ac)) / 2a
Plugging in the values from our equation:
a = -0.7
b = 35.0
c = -175
time = (-35.0 ± sqrt(35.0^2 - 4 * (-0.7) * (-175))) / (2 * -0.7)
Calculating further, we find:
time = (-35.0 ± sqrt(1225 + 490)) / -1.4
time = (-35.0 ± sqrt(1715)) / -1.4
Since the square root of 1715 is approximately 41.41, we get two values:
time1 = (-35.0 + 41.41) / -1.4 ≈ -4.58
time2 = (-35.0 - 41.41) / -1.4 ≈ 50.65
Since time cannot be negative, we discard time1 and keep time2.
The time it takes for Sue's car to reach the van is approximately 50.65 seconds.
Now let's calculate how far into the tunnel the collision occurs:
Using the equation for distance traveled:
distance = initial velocity * time + 0.5 * acceleration * time^2
For Sue's car:
initial velocity = 35.0 m/s
acceleration = -1.40 m/s^2
time = 50.65 s
distance = 35.0 * 50.65 + 0.5 * (-1.40) * 50.65^2
Simplifying the equation, we get:
distance ≈ 1750.78 m
Therefore, the collision occurs approximately 1750.78 meters into the tunnel.
To determine whether there will be a collision between Speedy Sue and the slow-moving van, we need to compare the distance it takes for Sue to stop or avoid the van with the distance between them.
First, let's find out how far Sue can brake when she applies the brakes. To do this, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
vf is the final velocity (0 m/s since Sue wants to stop),
vi is the initial velocity of Sue's car (35.0 m/s),
a is the acceleration (in this case, -1.40 m/s^2), and
d is the distance Sue is able to brake.
Plugging in the values, we get:
0 = (35.0 m/s)^2 + 2(-1.40 m/s^2)d
Simplifying the equation, we find:
0 = 1225 m^2/s^2 - 2.8 m/s^2d
Solving for d, we get:
2.8 m/s^2d = 1225 m^2/s^2
d = 1225 m^2/s^2 / 2.8 m/s^2
d ≈ 437.5 m
So Sue can brake for approximately 437.5 meters before coming to a stop.
Now, let's calculate the distance between Sue's car and the van. Since they are moving towards each other in the tunnel, their respective distances from the tunnel entrance can be added together.
The distance between them is:
Distance = (175 m + 437.5 m)
Distance ≈ 612.5 m
Now, let's find out the time it takes for Sue's car to reach the van. We can use the equation:
vf = vi + at
Where:
vf is the final velocity (0 m/s, since Sue wants to stop),
vi is the initial velocity of Sue's car (35.0 m/s),
a is the acceleration (in this case, -1.40 m/s^2), and
t is the time taken.
Plugging in the values, we get:
0 = 35.0 m/s + (-1.40 m/s^2)t
Simplifying the equation, we find:
1.40 m/s^2t = 35.0 m/s
t = 35.0 m/s / 1.40 m/s^2
t ≈ 25.0 s
Therefore, the collision occurs after approximately 25.0 seconds into the tunnel, at a distance of 612.5 meters from the tunnel entrance.