Answer:
there is a problem in question is hafe only
We need more detail I will help when i get a real question
Answer:
Step-by-step explanation:
Given that acceleration of an object is
is the solution to the differential equation
Since v(0) =7
we get ln 7 = C
Hence 
since velocity is rate of change of distance s we have
![v=\frac{ds}{dt} =7e^{-2t}\\s= [tex]s(t) =\frac{-7}{2} (e^{-2t})+C)[](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bds%7D%7Bdt%7D%20%3D7e%5E%7B-2t%7D%5C%5Cs%3D%20%5Btex%5Ds%28t%29%20%3D%5Cfrac%7B-7%7D%7B2%7D%20%28e%5E%7B-2t%7D%29%2BC%29%5B)
substitute t=0 and s=0
So solution for distance is
Answer:
the closest part the mechanic can chose is 0.5 inch
Step-by-step explanation:
fractional sizes given to the nearest 64th of an inch = 1/64 = 0.0156
the closest size to 0.492in = 0.492 + 0.0156 = 0.5076 inches
approximately, the closest part the mechanic can chose is 0.5 inch
We know that
speed=distance/time
solve for time
time=distance/speed
in this problem
<span>Marco runs at a rate of 6 miles per hour.
</span><span>Fernando funs at a rate of 7.2 miles per hour
Difference=7.2-6=1.2 miles/hour
so
speed=1.2 miles/hour
distance=0.3 miles
time=?
</span>time=distance/speed-----> 0.3/1.2-----> 0.25 hour-----> 0.25*60=15 minutes
<span>
the answer is
0.25 hour (15 minutes)
Alternative Method
Let
x---------> Fernando's distance when Marco is 0.3 miles apart
</span>Fernando funs at a rate of 7.2 miles per hour
<span>for distance =x
time=x/7.2------> equation 1
</span>Marco runs at a rate of 6 miles per hour.
for distance=x-0.30
time=(x-0.30)/6------> equation 2
equate equation 1 and equation 2
7.2*(x-0.3)=6x-----> 7.2x-2.16=6x
7.2x-6x=2.16------> x=2.16/1.2-------> x=1.8 miles
time=x/7.2-----1.8/7.2=0.25 hour
