Speedy Sue during is driving at 30.0m/s

and enters a 1 lane tunnel. She then observes a slow moving van 155m ahead traveling at 5.0m/s. She applies the brakes but can only accelerate at -2.0m/s^2 b/c road is wet.

a) will there be a collision? How do you know?
b) If there is a collision state how far in the tunnel and at what time the collision occurs. If not then determine the distance of closest approach btwn the car and van.

~well I know that
sue:
vi= 30.0m/s
a= -2.0m/s^2
xi= 155m

Van:
vi= 5.0m/s

I'm not sure what equation or what to do next. Is the acceleration constant? I assume so since this chap is about kinematics but it only says -2.0m/s^2 for acceleration so I guess it is..but do I need to find the xf for sue?

I think I would need to find xf but I don't have the final vf for Sue's car. Do I need to find that?

I need help on this..

Thank you

since v= d/t

would the time it takes to reach the van be
155m/30.0m/s= 5.17s?

~I don't think this is okay though since the van is moving so this just confuses me further..

To determine if there will be a collision, we need to calculate the distance it takes for Sue to stop (xf) and compare it to the distance the van is currently at (xi).

First, let's find the time it takes for Sue to stop. We can use the equation:
vf = vi + at

where:
vf is the final velocity (0 m/s, since Sue stops)
vi is the initial velocity (30.0 m/s)
a is the acceleration (-2.0 m/s^2)
t is the time it takes to stop (unknown)

Rearranging the equation, we get:
t = (vf - vi) / a

Substituting the values:
t = (0 - 30.0) / (-2.0)
t = 15.0 seconds

Now, we can find the distance it takes for Sue to stop using the equation:
xf = vi*t + (1/2)*a*t^2

Substituting the values:
xf = 30.0 * (15.0) + (1/2) * (-2.0) * (15.0)^2
xf = 450.0 - 225.0
xf = 225.0 meters

Comparing the distance Sue needs to stop (xf) with the distance of the van (xi) to the entrance of the tunnel (155 meters):
225.0 meters > 155 meters

Since the distance Sue needs to stop is greater than the distance of the van, there will be a collision.

To find the location of the collision in the tunnel, we need to calculate the distance covered by Sue in the time it takes for her to stop (15.0 seconds):
x = xi + vi*t + (1/2)*a*t^2

Substituting the values:
x = 155 + 30.0 * (15.0) + (1/2) * (-2.0) * (15.0)^2
x = 155 + 450.0 - 225.0
x = 380 meters

Therefore, the collision will occur 380 meters into the tunnel, and it will occur at the time it takes Sue to stop, which is 15.0 seconds.

Since there is a collision, there is no need to determine the distance of closest approach between the car and van.

To solve this problem, you can use the equations of motion and the concept of relative velocity.

a) To determine if there will be a collision, you need to compare the distance traveled by Speedy Sue before she comes to a stop with the distance between her initial position and the van's position.

First, you need to find the time it takes for Speedy Sue to come to a stop. You can use the equation:

v_f = v_i + at

Where:
v_f is the final velocity, which is 0 since Speedy Sue comes to a stop.
v_i is the initial velocity of Speedy Sue, which is 30.0 m/s.
a is the acceleration, which is -2.0 m/s^2.

Rearranging the equation, you have:

0 = 30.0 m/s + (-2.0 m/s^2)t

Solving for t, you find:

t = 30.0 m/s / 2.0 m/s^2 = 15.0 s

Now, you can determine the distance traveled by Speedy Sue before coming to a stop using the equation:

s = v_i*t + (1/2)a*t^2

Where:
s is the distance traveled by Speedy Sue.
v_i is the initial velocity of Speedy Sue, which is 30.0 m/s.
t is the time it takes for Speedy Sue to come to a stop, which is 15.0 s.
a is the acceleration, which is -2.0 m/s^2.

Plugging in the values, you get:

s = (30.0 m/s)*(15.0 s) + (1/2)*(-2.0 m/s^2)*(15.0 s)^2
s = 450.0 m - 225.0 m
s = 225.0 m

Now, you can compare this distance to the initial distance between Speedy Sue and the van, which is 155 m. Since the distance traveled by Speedy Sue before coming to a stop (225 m) is greater than the initial separation distance (155 m), there will be a collision.

b) To determine the distance and time of the collision, you need to find the time at which Speedy Sue catches up with the van.

The relative velocity between Speedy Sue and the van can be found by subtracting the van's velocity from Speedy Sue's velocity:

v_r = v_sue - v_van
v_r = 30.0 m/s - 5.0 m/s
v_r = 25.0 m/s

Now, you can determine the time it takes for Speedy Sue to catch up with the van using the equation:

t = (s_r) / (v_r)

Where:
t is the time it takes for Speedy Sue to catch up with the van.
s_r is the initial separation distance between Speedy Sue and the van, which is 155 m.
v_r is the relative velocity between Speedy Sue and the van, which is 25.0 m/s.

Plugging in the values, you get:

t = 155 m / 25.0 m/s
t ≈ 6.2 s

Now, you can determine the distance at which the collision occurs by multiplying the relative velocity by the time:

d = v_r*t
d = 25.0 m/s * 6.2 s
d ≈ 155.0 m

Therefore, the collision occurs when Speedy Sue is approximately 155.0 m into the tunnel, after 6.2 s.

If there is no collision, you can determine the distance of closest approach by using the equations of motion and finding the minimum distance between Speedy Sue and the van.

The acceleration is assumed to be constant until the car has a velocity of zero w.r.t. the ground or when it collides with the van.

You could write down coordinates of the car and the van as a function of time and see if at some time the become equal.

A simpler way to solve this problem is as follows. Suppose you are in the van and look at the car approaching you, what do you see?

What you see is a car initially (at t = 0) at a distance of 155 m that is approaching you at t= 0 at a speed of 25 m/s, and acceleating at -2 m/s^2. The car will either collide with the van or it will not reach the van. The braking will then stop when the car is moving away from the van at a speed of 5 m/s.

This is similar to someone on a tower watching someone throwing a ball upward in his direction. Will the ball reach him? The ball has some initial velocity in the upward direction, but is accelerating in the downward direction. Of course the ball will continue to accelerate downward and not stop to accelerate toward the ground once a certain speed is reached, but that is irrelevant to the question of whether the ball will reach a certain height.

In case of throwing a ball, you know that you can solve that using conservation of energy. The initial kinetic energy of the ball is
1/2 m v^2 where v is the initial speed. At it's highest point the ball will be at rest, therefore the height it will reach is:

m g h = 1/2 m v^2 --->

h = 1/2 v^2/g

In case of the car-van problem v = 25 m/s and g is 2 m/s^2. Inserting these numbers gives:

h = 156.25 m

This means that the car will hit the van.