Anna is driving from champaign to indianapolis on i-74. she passes the prospect ave. exit at noon and maintains a constant speed of 75 mph for the entire trip. chuck is driving in the opposite direction. he passes the brownsburg, in exit at 12: 30pm and maintains a constant speed of 65 mph all the way to champaign. assume that the brownsburg and prospect exits are 105 miles apart, and that the road is straight. how far from the prospect ave. exit do anna and chuck pass each other?

73.5 miles from Prospect Ave. exit.

Explanation:

From the question information, we know these two important facts:

1) Both drivers travel in a straight road.

2) Both drivers travel at a constant speed, although in opposite directions.

With this information, applying the definition of velocity (change in displacement divided by  change in time), we can write an equation for final displacement, applying it to each driver, as follows:

x = x₀ + v. (t- t₀)

Let´s take Anna first, naming her as driver 1.

As we want to know the distance from her starting point, when they pass each other, we choose  to define the initial parameters as follows:

x₁₀ = 0 and t₁₀ = 0

So, calling x1 to Anna's displacement, we can write:

x₁= 75 mph. t₁

For the second driver, as he starting from a point at 105 miles from the origin, at a time 0.5 h delayed regarding t₁, and is travelling in the opposite direction, the equation for Chuck's  displacement (that we will call x₂) is as follows:

x₂ = 105 m -65 (t₁-0.5 h)

Now, at the moment that they pass each other, both displacements will be the same: x₁= x₂.

So, we will have the following equation:

75 mph . t₁ = 105 m - 65. (t₁-0.5 h)

Solving for t₁, we get :

140 t₁ = 137.5 ⇒ t₁ = 0.98 h.

Replacing in the expression for x₁, we get:

x₁ = 75 mph. 0.98 h = 73.5 miles.

So, both drivers will pass each other at 73.5 miles from Prospect Ave. exit.

where’s the list ? the question is incomplete

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s