so, at time t hours, we have the distance z between the vehicles is
z^2 = (50-160t)^2 + (40+140t)^2
2z dz/dt = -2*160(50-160t) + 2*140(40+140t)
dz/dt = 90400t - 4800
that is in km/hr, so you need to figure 12 min = 1/5 hour
A police car traveling south toward Sioux Falls, Iowa, at 160km/h pursues a truck traveling east away from Sioux Falls at 140km/h.
An illustration shows a police car and a truck, both are moving towards Sioux Falls. The interconnecting path shows a right angled triangle. The car is moving downward at 160 kilometers per hour toward Sioux Falls which is at a distance y. A truck is moving at 140 kilometers per hour at a distance of x, away from Sioux Falls.
At time 𝑡=0, the police car is 50km north and the truck is 40km east of Sioux Falls.
Calculate the rate at which the distance between the vehicles is changing at 𝑡=12 minutes.
(Use decimal notation. Give your answer to three decimal places.)
rate:
2 answers
This work is correct up until “ dz/dt = 90400t - 4800”
If you look in the previous step, you see that 2z dz/dt is equal to this, so the mistake is in the fact that the answer was no divided by 2z
The correction is (90400t-4800)/2z
You then solve for z using the pythagorean theorem.
y will equal 50-160t and x will equal 40+140t
Plug in time for all values, then go back and use the corrected formula.
If you look in the previous step, you see that 2z dz/dt is equal to this, so the mistake is in the fact that the answer was no divided by 2z
The correction is (90400t-4800)/2z
You then solve for z using the pythagorean theorem.
y will equal 50-160t and x will equal 40+140t
Plug in time for all values, then go back and use the corrected formula.