Ahonda civic is travelling westbound on taylor st. at a constant speed of 20 miles per hour. another car, 5 miles west of the honda, is traveling eastbound on taylor st. at a constant speed of 30 miles per hour. how long (in minutes) will it take for the cars to be 25 miles apart? provide a numerical answer without units.

It will take 36 min for the cars to be 25 miles apart

Explanation:

Hi there!

The position of the cars after a time "t" is calculated as follows:

x = x0 + v · t

Where:

x = position of the car at time t.

x0 = initial position.

v = velocity.

t = time.

Let´s consider the origin of the system of reference as the point at which the Honda is located initially. Let´s also consider the west direction as positive:

The position of the Honda at time t will be:

x = x0 + v · t

x = 0 + 20 mi/h · t = 20 mi/h · t

The position of the other car is calculated as follows:

x = 5 mi - 30 mi/h · t

(Notice that the velocity is negative because the car travels eastwards)

When the cars are 25 miles apart, the difference between their positions is 25 miles:

position of the Honda - position of the other car = 25 mi

(notice that for the difference to be positive, the position of the Honda (which is positive because it travels westbound) has to be first in the equation)

20 mi/h · t - (5 mi - 30 mi/h · t) = 25 mi

Solving for t

20 mi/h · t - 5 mi + 30 mi/h · t = 25 mi

50 mi/h · t = 30 mi

t = 30 mi / 50 mi/h

t = 0.6 h (60 min/1 h) = 36 min

It will take 36 min for the cars to be 25 miles apart.

the answer is definitely c.

prolly a beacuse the momentuem the car has increases while going down hill