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3 years ago

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równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma

postać v(t)=25-5t. Korzystając z rachunku całkowego wyznacz drogę hamowania autokaru.

1 answer:

Rasek [7]3 years ago

3 0

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.

$s =\int v(t) dt\\\\s =\int (25 - 5t)dt\\\\s= 25 t - 2.5 t^2 \\$

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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You want to make either the length or width variable (I chose the width variable to be by itself. It doesn't matter) by itself by:

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$2's$ $cancel$ $out$
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Subtract both sides by either length or width depending on what variable you chose to make by itself (in this case I subtracted the length variable because I wanted the width variable by itself).
$\frac{p}{2} - l = w + l - l$
$l's$ $cancel$ $out$
p/2 - l = w

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You replace w with the equation above p/2 - l = w
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Multiply out the L.
$a = \frac{p}{2}l - l^{2}$

Plug in all your numbers a = 110 and p = 42
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Move everything to the left side so $l^{2}$ would be positive (makes the equation easier when $l^{2}$ is positive).
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Factor.
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Make each parenthesis set equal to 0.
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Add.
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