Suppose two cars depart from a four-way intersection at the same time, one heading north and the other heading west. The car heading north travels at the steady speed of 20 ft/sec and the car heading west travels at the steady speed of 36 ft/sec.
(a) Find an expression for the distance between the two cars after t seconds. (Round your coefficients to one decimal place as needed.) ft
(b) Find the distance in miles between the two cars after 3 hours 15 minutes. (Round your answer to one decimal place.) in mi
(c) When are the two cars 1 mile apart? (Round your answer to one decimal place.) in sec
this is what I have so far
(a) d=√((20t)^2+(36t)^2)=4√106t
(b) t=3*3600+15*60=11,700 in (a) / 5280
(c) solve for t when d = 5280
I get lost from here
a) correct, where d is in feet and t is in seconds, careful with brackets
d = (4√106)t
b) Your time is in hours, so 3 hours and 15 min = 3.25 hours
= 3.25*3600 s
= 11700 seconds
d = 4√(106) * 11700) = appr 481,835.49 ft
= 481,835.49/5280 miles
= appr 91.3 miles
c) when is 4√106 t = 1 mile = 5280 ft
4√106 t = 5280
√106 t = 1320
t = 1320/√106 = 128.2 seconds
= appr 2 minutes.
this makes sense, since 20 ft/s = (20/5280) / (1/3600) mph = appr 13.6 mph
and 36 ft/s = 24.5 mph (rounded off to 1 decimal)
so after 1 hours D = √(13.6^2 + 24.5^2) = 28 mile
so in 3.25 hours, D = 3.25(828) = 91
As a matter of fact it might have been easier to change it all to mph at the
beginning, but hey, what difference does it make with calculators.
What would you get for part (a) iam still lost on that one
a) you had d = 4√106 t, which I said is correct
so I don't understand why you are lost.
If you wanted the actual value of 4√106 that would be appr 41.2
but we should not round off too soon. With todays calculators there is no
need to round off until you reach the final answer.
I have a 25-year old SHARP, with 7 different memory locations.
(very sophisticated for those times)
I use these to store my intermediate answers and only after the last
calculation do I round off.
Well the answers needs to be in ft and it says 4√106 and 41.2 is wrong so ?
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sorry wrong website
To find the distance between two cars after a certain time, we can use the Pythagorean theorem. Let's label the distance between the two cars as "d" at time "t".
(a) We have already established that the car heading north travels at 20 ft/sec and the car heading west travels at 36 ft/sec. After time "t", the distance traveled by the northbound car would be 20t ft, and the distance traveled by the westbound car would be 36t ft.
Using the Pythagorean theorem, the distance "d" between the two cars can be found using the formula:
d = √((20t)^2 + (36t)^2)
Simplifying this equation, we get:
d = √(400t^2 + 1296t^2)
d = √(1696t^2)
d = √(1696) * √(t^2)
d = √(16 * 106) * t
d = 4√106t
So, the expression for the distance between the two cars after "t" seconds is 4√106t ft.
(b) To find the distance between the two cars after 3 hours and 15 minutes (or 3.25 hours), we need to substitute the value of "t" into our expression. Since we are asked to find the distance in miles, we also need to convert the result from feet to miles.
d = 4√106t
d = 4√106 * (3.25 * 3600) / 5280
d ≈ 295.5 ft (rounded to one decimal place)
To convert feet to miles, divide the distance by 5280:
d = 295.5 / 5280 ≈ 0.056 miles (rounded to one decimal place)
So, the distance between the two cars after 3 hours and 15 minutes is approximately 0.056 miles.
(c) To find when the two cars are 1 mile apart, we need to solve the equation:
d = 4√106t
1 = 4√106t
Squaring both sides and isolating "t", we have:
1 = 16 * 106 * t^2
t^2 = 1 / (16 * 106)
t ≈ √(1 / (16 * 106))
Calculating this value using a calculator or software, we get:
t ≈ 0.032 seconds (rounded to one decimal place)
So, the two cars are approximately 1 mile apart after 0.032 seconds.