A car is moving at 25.0m/s on a straight highway lane. 10.0m in front of the car is a van that is travelling with the same speed. The van’s driver suddenly hits the brakes to avoid an obstruction, slowing down at a rate of 9.00m/s². The car’s driver reacts 0.85s later, when he realizes the van’s tail lights are on. Assume that the brakes are the same quality and the car decelerates at the same constant rate as the car.
a. Calculate how far the van is going to travel before it stops.
b. Calculate how much distance the car needs to come to a stop.
c. Is the car going to rear-end the van?
d. If the car’s driver is distracted because of his phone (!) and reacts 1.20s later, instead of the 0.85s, how much distance will elapse before they collide?
e. How fast is the car going just as the collision happens, in the case described by part (d)?
lmfao the professor at uncc checks this bro
im guessing you have the answer then?
ughh just give me the equations i need to solve the problem pleasee
a. Calculate how far the van is going to travel before it stops.
***vf^2=vo^2 + 2a d. solve for d.
b. Calculate how much distance the car needs to come to a stop.
vf^2=vo^2 + 2ad. Now to this d, add vo*.85
c. Is the car going to rear-end the van?***does the car go 10 m further than the van? If so, it hits it.
d. If the car’s driver is distracted because of his phone (!) and reacts 1.20s later, instead of the 0.85s, how much distance will elapse before they collide?***same equations.
e. How fast is the car going just as the collision happens, in the case described by part (d)? set the distance = 10+van distance, and then solve for Vf
so for part b, it's the same answer as a, but i add (25m/s + 0.85s)?
To answer these questions, we will need to use the equations of motion. The key equation we will use is:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Let's solve each part of the problem step by step:
a. To calculate how far the van will travel before it stops, we need to find the time it takes for the van to come to a stop. We can use the equation:
v = u + at
where:
v = final velocity (0 m/s, since the van comes to a stop)
u = initial velocity (25.0 m/s)
a = acceleration (-9.00 m/s², since the van is decelerating)
t = unknown
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values:
t = (0 - 25.0) / (-9.00) = 2.7778 s
Now we can calculate the distance using the equation:
s = ut + (1/2)at²
where:
s = distance
u = initial velocity (25.0 m/s)
t = time (2.7778 s)
a = acceleration (-9.00 m/s²)
Substituting the values:
s = (25.0)(2.7778) + (1/2)(-9.00)(2.7778)²
s ≈ 69.4 m
Therefore, the van will travel approximately 69.4 meters before it stops.
b. To calculate how much distance the car needs to come to a stop, we can use the same approach as in part a, but with different values. The car's initial velocity is also 25.0 m/s, and the acceleration is also -9.00 m/s².
Using the same equation:
v = u + at
where:
v = final velocity (0 m/s, since the car comes to a stop)
u = initial velocity (25.0 m/s)
a = acceleration (-9.00 m/s²)
t = unknown
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values:
t = (0 - 25.0) / (-9.00) = 2.7778 s
Now, to find the distance, we use the equation:
s = ut + (1/2)at²
where:
s = distance
u = initial velocity (25.0 m/s)
t = time (2.7778 s)
a = acceleration (-9.00 m/s²)
Substituting the values:
s = (25.0)(2.7778) + (1/2)(-9.00)(2.7778)²
s ≈ 69.4 m
Therefore, the car needs to travel approximately 69.4 meters to come to a stop.
c. To determine whether the car will rear-end the van, we need to compare the distance the car needs to stop (69.4 m) with the distance between the car and the van at the moment the car's driver reacts (10.0 m).
Since the car needs more distance to stop than the distance between them, it will rear-end the van. In this scenario, the car will not have enough time and distance to stop before colliding with the van.
d. If the car's driver is distracted and reacts 1.20 s later instead of the original 0.85 s, we need to calculate how much distance elapses before they collide.
First, we need to calculate the total time from the moment the van stops to the moment of collision:
Total time = time for the car to react + time for the van to stop
Time for the car to react = 1.20 s (given)
Time for the van to stop = 2.7778 s (from part a)
Total time = 1.20 s + 2.7778 s ≈ 3.9778 s
To find the distance traveled by the van during this time, we use the equation:
s = ut + (1/2)at²
where:
s = distance
u = initial velocity (25.0 m/s)
t = time (3.9778 s)
a = acceleration (-9.00 m/s²)
Substituting the values:
s = (25.0)(3.9778) + (1/2)(-9.00)(3.9778)²
s ≈ 79.5 m
Therefore, approximately 79.5 meters will elapse before the car and van collide.
e. To determine the speed of the car just as the collision happens in the scenario described in part d, we need to calculate the final velocity of the car.
Using the same equation:
v = u + at
where:
v = final velocity
u = initial velocity (25.0 m/s)
a = acceleration (-9.00 m/s²)
t = time (3.9778 s)
Substituting the values:
v = 25.0 - 9.00(3.9778)
v ≈ -4.5 m/s
The negative sign indicates that the car is moving in the opposite direction of the initial motion. Therefore, the car's speed just as the collision happens is approximately 4.5 m/s in the opposite direction.
Please note that in real-world situations, it is important to prioritize safety and avoid distracted driving.