# why is the integral used in this question? (1 Viewer)

#### RakeshCristoval

##### New Member
and also, how do you know whether to use integrals or not in questions like these?

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#### TheShy

##### Member
If you integrate an acceleration-time graph, you get the velocity.
Something to note : Differentiating displacement/distance-time graph gives velocity. Integrating gives nothing
differentiating velocity-time graph gives displacement/distance. Integrating gives acceleration
Differentiating acceleration-time graph gives nothing, Integrating gives velocity

#### RakeshCristoval

##### New Member
forgot what natural logs are. the answer starts off by integrating it, idk why

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#### Velocifire

##### Well-Known Member
Sorry, just that I was scared people would say I'm waaaaaaaay off.

#### RakeshCristoval

##### New Member
If you integrate an acceleration-time graph, you get the velocity.
Something to note : Differentiating displacement/distance-time graph gives velocity. Integrating gives nothing
differentiating velocity-time graph gives displacement/distance. Integrating gives acceleration
Differentiating acceleration-time graph gives nothing, Integrating gives velocity

#### Velocifire

##### Well-Known Member
If you integrate an acceleration-time graph, you get the velocity.
Something to note : Differentiating displacement/distance-time graph gives velocity. Integrating gives nothing
differentiating velocity-time graph gives displacement/distance. Integrating gives acceleration
Differentiating acceleration-time graph gives nothing, Integrating gives velocity
Oh yeah, like physics, finding the slope and area to find acc jerk and other essential pieces of data but those graphs were straight lines.

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#### Drdusk

##### π
Moderator
and also, how do you know whether to use integrals or not in questions like these?

Adding onto TheShy, a = dv/dt so naturally you need to integrate both sides to get v in terms of t because v and t are what you get when you integrate dv and dt respectively.

#### RakeshCristoval

##### New Member
Adding onto TheShy, a = dv/dt so naturally you need to integrate both sides to get v in terms of t because v and t are what you get when you integrate dv and dt respectively.
right okay, thank you

#### mikikieko12

##### New Member
You need to integrate this mainly because of the projectile motion math (forgot the name of the topic...I think rates of change or somethin).

For this specific question, here's one method to look at it, and a way to identify whether to integrate or differentiate:

In terms of differentiation:
Distance ---> (first derivative) = velocity ---> (second derivative) = acceleration

Integral is simply the reverse:
Acceleration ---> first integral = velocity ---> 2nd integral = distance

Also you need to make sure to analyse the question first! Look for what the question is asking, and logically answer it. To know which questions to integrate you need to practice numerous questions and also learn specific terminology (especially projectile motion ones) as it always hints whether to integrate or differentiate.

Hope this helps ;;;

#### CM_Tutor

##### Well-Known Member
If you integrate an acceleration-time graph, you get the velocity.
Something to note : Differentiating displacement/distance-time graph gives velocity. Integrating gives nothing
differentiating velocity-time graph gives displacement/distance. Integrating gives acceleration
Differentiating acceleration-time graph gives nothing, Integrating gives velocity
Two points:

First, there is a typo above. The line "differentiating velocity-time graph gives displacement/distance. Integrating gives acceleration" is backwards, it should read that "differentiating velocity-time graph gives acceleration. Integrating gives displacement."

Second, be careful not to confuse distance and displacement. Displacement or position is given by x and it has a direction, distance does not.

Example: Suppose a particle is moving according to $\bg_white a = 2t - 1$ and it starts from rest at $\bg_white x = 2$. Find the distance it travels in the first 2 seconds of motion.

\bg_white \begin{align*} a &= 2t - 1 \\ \text{Since,} \quad a &= \cfrac{dv}{dt} \\ \cfrac{dv}{dt} &= 2t - 1 \\ v &= \int 2t -1 \; dt = \cfrac{2t^2}{2} - t + C_1 \text{, for some constant C_1} \\ \text{Now, at t = 0, v = 0}: \quad 0 &= 0^2 - 0 + C_1 \implies C_1 = 0 \\ \text{So,} \quad v &= t^2 - t \\ \\ \text{Now, we also know that} \quad v &= \cfrac{dx}{dt} \\ \text{and so} \quad \cfrac{dx}{dt} &= t^2 - t \\ x &= \int t^2 - t \; dt = \cfrac{t^3}{3} - \cfrac{t^2}{2} + C_2 \text{, for some constant C_2} \\ \text{Now, at t = 0, x = 2}: \quad 2 &= \cfrac{0^3}{3} - \cfrac{0^2}{2} + C_2 \implies C_2 = 2 \\ \text{So,} \quad x &= \cfrac{t^3}{3} - \cfrac{t^2}{2} + 2 \\ \\ \text{Now, at t=2}: \quad x &= \cfrac{2^3}{3} - \cfrac{2^2}{2} + 2 = \cfrac{8}{3} - 2 + 2 = +2\cfrac{2}{3} \end{align*}

So, we know that the object starts at rest at $\bg_white x = 2$ and after 2 seconds, reaches $\bg_white x = 2\frac{2}{3}$. It's displacement over this period is thus $\bg_white \frac{2}{3}$ of a unit in the positive direction from its starting point, but does this mean that the distance it has traveled is $\bg_white \frac{2}{3}$? Not necessarily. If I get up from my desk, walk to the kitchen to get a drink, and then walk into the lounge room, my displacement is the change in position from desk to lounge room, but my distance traveled is further. For a distance calculation, I need to check if the object stops anywhere (as direction can only change when $\bg_white v = 0$:

\bg_white \begin{align*} \text{Put} \quad v &= 0 \\ t^2 - t &= 0 \\ t(t - 1) &= 0 \end{align*}

We knew from the question that the particle was at rest at $\bg_white t = 0$, but it is now clear that it is also at rest at $\bg_white t = 1$.

\bg_white \begin{align*} \text{Put t = 1 into x} \quad x &= \cfrac{t^3}{3} - \cfrac{t^2}{2} + 2 \\ &= \cfrac{1^3}{3} - \cfrac{1^2}{2} + 2 \\ &= \cfrac{1}{3} - \cfrac{1}{2} + 2 \\ &= \cfrac{2}{6} - \cfrac{3}{6} + 2 \\ &= 1\cfrac{5}{6} \end{align*}

So, the particle starts at $\bg_white x = 2$ (at $\bg_white t = 0$), moves to $\bg_white x = 1\frac{5}{6}$ (at $\bg_white t = 1$) where it changes direction (you can check by seeing that for $\bg_white 0 < t < 1$, $\bg_white v < 0$ and after $\bg_white t > 1$, $\bg_white v > 0$) reaching $\bg_white x = 2\frac{2}{3}$ by $\bg_white t = 2$. It follows that the distance it travels in those 2 seconds is:

\bg_white \begin{align*} \text{Total distance } &= \text{ Distance traveled from t = 0 to t = 1 } + \text{ Distance traveled from t = 1 to t = 2} \\ &= \bigg(2 - 1\cfrac{5}{6}\bigg) + \bigg(2\cfrac{2}{3} - 1\cfrac{5}{6}\bigg) \\ &= (2 - 1) + \bigg(0 - \cfrac{5}{6}\bigg) + (2 - 1) + \bigg(\cfrac{2}{3} - \cfrac{5}{6}\bigg) \\ &=\bigg(1 - \cfrac{5}{6}\bigg) + 1 + \bigg(\cfrac{4}{6} - \cfrac{5}{6}\bigg) \\ &= \cfrac{1}{6} + 1 - \cfrac{1}{6} \\ &= 1 \end{align*}

Similar care needs to be taken in considering speed as opposed to velocity, or their averages.