Distance Covered by a Car The velocity of the Zephyr electric sedan traveling along a straight r

step 1 Welcome to this lesson. In this lesson we have the velocity and you're looking for distance. So

Distance Covered by a Car The velocity of the Zephyr...

Key Concepts

Velocity and Displacement

This concept explains that the displacement (or distance traveled) of an object is the accumulated effect of its velocity over time. In calculus, displacement is obtained by integrating the velocity function over a given interval, which is a key idea in kinematics.

Definite Integration

Definite integration is the process of computing the area under a curve within a specified interval. It is used to determine the total change, such as the total distance covered by an object when its velocity function is integrated over time.

Fundamental Theorem of Calculus

This theorem links the concepts of differentiation and integration, showing how the integral of a function’s derivative over an interval yields the net change in the function over that interval. It is essential in understanding how integrating the velocity function results in the overall displacement of the object.

Transcript

00:01 Welcome to this lesson.

00:02 In this lesson we have the velocity and you're looking for distance.

00:14 So we'll take the integral, this from 0 to 6, it's called to the integral t plus 1, the t from 0 to 6.

00:35 So for this one we'll do substitution.

00:42 And if you're doing that the bounce, we're changing to the one in seven also would add one to zero and six so the u would be equal to d t and also t would be called to u minus one so that we can make all the changes so 0 to 6 8 t land t plus 1 the t now becomes 8 we have u you minus 1 okay so we can expand this is learn u to you so let's span we have 1 to 7 then 8 u then u minus 8 learn u so i so i can let all of them be on the arrow.

02:56 Okay, so with this one, we'll solve it one after the other.

03:00 Let's take only this.

03:11 So with this one, we would introduce f of u and j prime of you, and this would be f of u, j of u, so this one will let f of u equals to 8 u we'll let it equal to learn u that f prime of u becomes 1 over u and also will let the prime of u be equal to 8 u so that u becomes 4 u squared that is if you take the differentiation both sides so i can write h u then u d u as lean u then j of u so this one comes first minus j of u which is 4 u times f prime of u which is one over u so this you can go away and we would introduce the bounce so this is equal to four times 44 times 49 that is one six one nine six main u so only u that's nine seven then minus if i put one there i have four lane one and one is zero so with this one, two, i have two, u squared.

06:01 And that is one, seven.

06:04 So this becomes 196, length, seven, minus.

06:11 Now we have 47, 49 times two, which is 98, then minus, if i put one here, i have two.

06:28 So 98 minus two, which is 96.

06:35 Okay so this is what i have for 8 u then you the last part is this part where i have 8 then you do you okay so i have 8 then you and you so this one now use the i use the u you the you the way approach you the approach and this is equal to uv minus the integral of a d u so with this one now let's learn u because u so that one of uh that causes you have to replace it with s and t so x d t so that like i have x t minus integral t d s okay so with this one i would have x that is equal to lane u i have bx is equal to 1 over u okay and i also have the d u that is equal to d t all right and this implies that that t is equal to u okay so if i substitute i have s which is then u i have t which is u so u lenu i will not forget the bounce that is one and seven so minus t and now you have ds which is d u d u yes is actually one over u, so that this cancels that.

09:48 And i have only du here...

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