Speedy sue, driving at 30.0 m/s, enters a one lane tunnel. she then observes a slow moving van 155 m ahead traveling at 5.00 m/s. sue applies her brakes, but can accelerate only at -2.00 m/s2 because the road is wet. will there be a collision? state how you decide. if yes, determine how far into the tunnel and what time the collision occurs. if no, determine the distance of closest approach between sue’s car and the van.

There will be a collision 11.4 s after Sue observes the van and hit the brakes.

The collision will occur at 212 m inside the tunnel.

Explanation:

The position and velocity of the Sue´s car can be calculated as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time at which Sue´s car and the van meet. At that time, their position will be the same. Let´s place the origin of the system of reference at the point where Sue starts decelerating so that the initial position for Sue is 0 m. Then:

Position of Sue´s car = position of the van

Sue´s car position:

x = x0 + v0 · t + 1/2 · a · t²

x = 0 m + 30.0 m/s · t - 1/2 · 2.00 m/s² · t²

Position of the van:

Since the van travels at constant velocity, a = 0. Then:

x = x0 + v · t

x = 155 m + 5.00 m/s · t

Position of Sue´s car = position of the van

30.0 m/s · t - 1/2 · 2 m/s² · t² = 155 m + 5.00 m/s · t

Let´s solve the equation for t:

30.0 m/s · t - 1.00 m/s² · t² = 155 m + 5.00 m/s · t

- 1.00 m/s² · t² + 30 m/s · t - 5.00 m/s · t - 155 m = 0

-1.00 m/s² · t² + 25 m/s · t - 155 m = 0

Solving the quadratic equation

t = 11.4 s and t = 13.6 s

Let´s consider the lower value because if the collision occurs at t = 11.4, it will not occur at a later time again.

Let´s calculate the velocity of Sue´s car at t = 11.4. If it is greater than zero, then there will be a collision.

v = v0 + a · t

v = 30.0 m/s - 2.00 m/s² · 11.4 s

v = 7.2 m/s

Then, there will be a collision 11.4 s after Sue observes the van and hit the brakes.

The distance inside the tunnel at which the collision occurs can be calculated using the equation of position of the van (or Sue´s car) at t = 11.4 s.

The position of the van at t = 11.4 s is calcualted as follows:

x = x0 + v · t

x = 155 m + 5.00 m/s · 11.4 s

x = 212 m

The collision will occur at 212 m inside the tunnel.

the rearrangement of particles in a physical change. hope this

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j