The falling object in Example 2 satisfies the initial value...

The falling object in Example 2 satisfies the initial value problem dv/dt = 9.8 - (v/5), v(0) = 0. (a) Find the time that must elapse for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)?

Transcript

00:01 Hello students in this question they are giving dv divided by d t equal to 9 .8 subtracting v divided by 5 and the initial velocity v of 0 is equal to 0 and we have to find time for object to reach 98 percentage of its velocity.

00:17 Let's start the sum dv divided by d t equal to 9 .8 subtracting v divided by 5.

00:26 Here we're taking lcm then 5 is here here 9 .8 multiply by 5 which is equal to 49.

00:36 Then we can write here 49 subtracting b.

00:41 Here dv divided by d t here by using variable separate method then we get 1 divided by 49 subtracting b and multiply by dv then equal to here 1 divided by 5 then d t is here multiply by d t now we can integrating on both side then integral 1 divided by 49 subtracting v and multiply by dv equal to integral 1 divided by 5 and multiply by d t okay we can solve this then we get negative natural log then 49 subtracting v equal to 1 divided by 5 multiply by t plus constant c.

01:39 And here we taking exponents on both side.

01:45 Then e power natural log 49 subtracting v and here e power negative of 1 divided by 5 t addition c.

01:57 Now we can cancel exponents and natural log then remind 19 subtracting v equal to we can write the constant c here then a power subtracting 1 divided by 5 t then we is equal to 49 and subtracting c e power 1 divided by 5 t here we substitute v is equal to 0 then initial velocity is equal to 49 subtracting c e the power 1 divided by 5 t is here 0 according to the sum we have 0 equal to 0 then we can substitute 0 equal to 49 and here anything power 0 equal to 0 then 49 subtracting c then c equal to 49 we can substitute the value of c in here then v of t equal to 49 subtracting 49 e power negative 1 divided by 5 times t it can be written as v of t equal to we can taking 49 as common then one subtracting a power minus 1 d by 5 times t so this is the equation of velocity so here we can find the time for object to reach 98 percentage of its limiting velocity here 98 percentage of 49 is equal to multiplied by 98 divided by 100 which is equal to 48 .02.

03:41 Now we can substitute the values in the equation then velocity of 90 percentage here 48 .02 equal to here 49 then one subtracting e power minus 1 divided by 5 times of t can be returned as minus t divided by 5 divided by 5...

Question

The falling object in Example 2 satisfies the initial value problem dv/dt = 9.8 - (v/5), v(0) = 0. (a) Find the time that must elapse for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)?

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