Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 175 m ahead traveling at 5.40 m/s. Sue applies her brakes but can accelerate only at −1.80 m/s2 because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs.
for Sue:
a = -1.8
v = -1.8t + c
when t = 0 , (she applied the brake), her v = 34 m/s
34 = -1.8(0) + c
c = 35.8
Sue's velocity equation: v = -1.8t + 35.8
Sue's distance equation:
d = -.9t^2 + 35.8t + k
let's look at the distance when she applied the break, that is, when t=0
d = 0 + 0 + k, so k = 0 and
d = -.9t^2 + 35.8t
after t seconds, the car in front of her will be (175 + 5.4t)
after t seconds Sue will have travelled -.9t^2 + 35.8t m
If there is a collision those must be equal, that is
-.9t^2 + 35.8t = 175 + 5.4t
-.9t^2 + 30.4t - 175 = 0
.9t^2 - 30.4t + 175 = 0
t = (30.4 ± √294.16)/1.8
= appr 7.138 or 26.4 , we will reject 26.4 since Sue has already collided
when t = 7.138
d = 175 + 5.4(7.138) = 213.5 m into the tunnel
check my arithmetic
To determine if there will be a collision, we can calculate the time it will take for Speedy Sue to reach the slow-moving van and see if it is less than the time it would take for Sue to stop.
We can use the following formula to calculate the time it would take for Sue to reach the slow-moving van:
time = (final velocity - initial velocity) / acceleration
For Sue:
- Initial velocity (u) = 34.0 m/s
- Final velocity (v) = 5.40 m/s
- Acceleration (a) = -1.80 m/s² (negative because it's deceleration due to braking)
Substituting the values in the formula:
time = (5.40 m/s - 34.0 m/s) / -1.80 m/s²
Calculating the time:
time = (-28.6 m/s) / (-1.80 m/s²)
time = 15.89 s
Now that we have the time it would take for Sue to reach the van, we can calculate the distance she would have traveled in that time:
distance = initial velocity * time + (1/2) * acceleration * time²
For Sue:
- Initial velocity (u) = 34.0 m/s
- Time (t) = 15.89 s
- Acceleration (a) = -1.80 m/s²
Substituting the values in the formula:
distance = (34.0 m/s) * (15.89 s) + (1/2) * (-1.80 m/s²) * (15.89 s)²
Calculating the distance:
distance = 540.26 m
The distance Sue would have traveled in 15.89 s is approximately 540.26 m. If this distance is less than or equal to the initial distance between Sue and the van (175 m), there will be a collision.
Since the distance traveled is larger than the initial distance, there will be a collision.
To determine when the collision occurs and how far into the tunnel, we can use the formula:
distance = initial distance + (initial velocity * time) + (1/2) * acceleration * time²
For Sue:
- Initial distance (d) = 175 m
- Initial velocity (u) = 34.0 m/s
- Time (t) = 15.89 s
- Acceleration (a) = -1.80 m/s²
Substituting the values in the formula:
175 m + (34.0 m/s) * (15.89 s) + (1/2) * (-1.80 m/s²) * (15.89 s)²
Calculating the final distance:
distance = 376.06 m
Therefore, the collision occurs at approximately 376.06 m into the tunnel after 15.89 s.
To determine if there will be a collision, we can compare the distances traveled by Speedy Sue and the slow-moving van during the time it takes for Sue to come to a stop.
Let's begin by finding out the time it takes for Speedy Sue to stop her car. We can use the following kinematic equation:
v = u + at
where
v = final velocity (0 m/s, as Sue comes to a stop)
u = initial velocity (34.0 m/s)
a = acceleration (-1.80 m/s²)
Rearranging the equation to solve for time (t):
t = (v - u) / a
Substituting the given values:
t = (0 - 34.0) / (-1.80)
t = 18.89 seconds
Now that we have the time it takes for Sue to stop, we can calculate the distance traveled by both vehicles during this time.
Distance traveled by Sue:
Using the equation:
s = ut + 0.5at²
where
s = distance traveled
u = initial velocity (34.0 m/s)
t = time (18.89 seconds)
a = acceleration (-1.80 m/s²)
s = (34.0 × 18.89) + 0.5 × (-1.80) × (18.89)²
s = 321.99 meters
Distance traveled by the slow-moving van:
To find the distance traveled by the van during the same time, we can use the equation:
s = ut
where
s = distance traveled
u = velocity of the van (5.40 m/s)
t = time (18.89 seconds)
s = 5.40 × 18.89
s = 102.09 meters
Now we can compare the distances traveled by Sue and the van during the time it takes for Sue to stop:
322 meters (Sue's distance) - 102 meters (van's distance) = 220 meters
Since Sue will travel 220 meters into the tunnel during the time it takes for her to stop, and the distance to the van is 175 meters, there will be a collision.
The collision will occur 220 meters into the tunnel from the entrance, and it will take approximately 18.89 seconds for the collision to occur.