równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma
1 answer:
Answer:
s = 22.5 m
Step-by-step explanation:
the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.
v (t) = 25 - 5 t
at t = 0 , v = 20 m/s
Let the distance is s.
Let at t = t, the v = 20
So,
20 = 25 - 5 t
t = 1 s
So, s = 25 x 1 - 2.5 x 1 = 22.5 m
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Time taken by jerry alone is 10.1 hours
Time taken by callie alone is 8.1 hours
<u>Solution:</u>
Given:- For a certain prospectus Callie can prepare it two hours faster than Terry can
Let the time taken by Terry be "a" hours
So, the time taken by Callie will be (a-2) hours
Hence, the efficiency of Callie and Terry per hour is
If they work together they can do the entire prospectus in five hours
On cross-multiplication we get,
On cross multiplication ,we get
<em><u>using quadratic formula:-</u></em>
If we take a = 0.9, then while calculating time taken by callie = a - 2 we will end up in negative value
Let us take a = 10.1
So time taken by jerry alone = a = 10.1 hours
Time taken by callie alone = a - 2 = 10.1 - 2 = 8.1 hours