Motion + Integration (1 Viewer)

Makro · Jun 30, 2009

So I didn't realise I didn't do this exercise and I left it for a week and it seems I've forgotten how to do half of it. Here is a scanned copy of the textbook page
I tried and failed on:
Q8, 9, 10, 12, 13

Did the rest, so yeah, help would be much appreciated.

kurt.physics · Jun 30, 2009

Makro said:
So I didn't realise I didn't do this exercise and I left it for a week and it seems I've forgotten how to do half of it. Here is a scanned copy of the textbook page
I tried and failed on:
Q8, 9, 10, 12, 13

Did the rest, so yeah, help would be much appreciated.

So listing important information usually helps:

$\ddot{x}= -9.8$

$t = 0, x = 2$

$t = 0, v = 4$

Using this information:

$v = \int \ddot{x}\,\, dt$

$\therefore v = \int -9.8 \,\, dt$

$=-9.8 \int dt$

$\therefore v= -9.8t + C$

$t = 0, v = 4$

ie

$-9.8 \times 0 + C = 4$

$\therefore C = 4$

$\therefore v = -9.8t + 4$

$x = \int v dt$

$\therefore x = \int \left(-9.8t + 4\right) dt$

$= -9.8 \times \frac{t^2}{2} + 4t + C$

$\therefore x = -4.9t^2 + 4t + C$

$t = 0, x = 2$

ie

$-4.9 \times 0 + 4 \times 0 + C = 2$

$\therefore C = 2$

$\therefore x = -4.9t^2 + 4t + 2$

kurt.physics · Jun 30, 2009

makro said:
so i didn't realise i didn't do this exercise and i left it for a week and it seems i've forgotten how to do half of it. here is a scanned copy of the textbook page
i tried and failed on:
Q8, 9, 10, 12, 13

did the rest, so yeah, help would be much appreciated.

q9)We have

$\ddot{x} = \left(3t + 1\right)^2$

$t = 0, v = 0$

$t = 0, x = -2$

(2 m to the left of x means that the displacement is -2 (as it is to the left))

So,

$v = \int \ddot{x} dt$

$\therefore v = \int \left(3t + 1\right)^2 dt$

$= \int \left(9t^2 + 6t + 1\right) dt$

$= 9 \cdot \frac{t^3}{3} + 6 \cdot \frac{t^2}{2} + t + C$

$\therefore v = 3t^3 + 3t^2 + t + C$

$t = 0, v = 0$

ie

$3 \cdot 0^3 + 3 \cdot 0^2 + 0 + C = 0$

$\therefore C = 0$

$x = \int v dt$

$= \int \left(3t^3 + 3t^2 + t\right) dt$

$= 3 \cdot \frac{t^4}{4} + 3 \cdot \frac{t^3}{3} + \frac{t^2}{2} + C$

$= \frac{3t^4}{4} + t^3 + \frac{t^2}{2} + C$

$t = 0, x = -2$

ie

$\frac{3\cdot 0^4}{4} + 0^3 + \frac{0^2}{2} + C = -2$

$\therefore C = -2$

$\therefore x = \frac{3t^4}{4} + t^3 + \frac{t^2}{2} - 2$

When $t = 4$ ,

$x = \frac{3\cdot 4^4}{4} + 4^3 + \frac{4^2}{2} - 2$

$= 262$

kurt.physics · Jun 30, 2009

Makro said:
So I didn't realise I didn't do this exercise and I left it for a week and it seems I've forgotten how to do half of it. Here is a scanned copy of the textbook page
I tried and failed on:
Q8, 9, 10, 12, 13

Did the rest, so yeah, help would be much appreciated.

10)

$v = e^t - 1$

$t = 0, x = 3$

(Because it is 3m to the right it is positive)

$x = \int v \,\, dt$

$\therefore x = \int \left(e^t - 1\right) \,\, dt$

$= e^t - t + C$

$t = 0, x = 3$

ie

$e^0 - 0 + C = 3$

$1 + C = 3$

$\therefore C = 2$

$\therefore x = e^t - t + 2$

when $t=5$ ,

$x = e^5 - 5 + 2$

$= \left(e^5 - 3\right) m$

kurt.physics · Jun 30, 2009

Makro said:
So I didn't realise I didn't do this exercise and I left it for a week and it seems I've forgotten how to do half of it. Here is a scanned copy of the textbook page
I tried and failed on:
Q8, 9, 10, 12, 13

Did the rest, so yeah, help would be much appreciated.

12)

$\ddot{x} = -9 \sin 3t$

$t = 0, v = 5$

$t = 0, x = -3$

$v = \int \ddot{x} \,\,dt$

$\therefore v = \int -9 \sin 3t \,\,dt$

$= -9 \int \sin 3t \,\,dt$

$= -9 \cdot - \frac{1}{3} \cos 3t + C$

$= 3 \cos 3t + C$

$t = 0, v = 5$

ie

$3 \cos\left(3 \cdot 0\right) + C = 5$

$3 \cos 0 + C = 5$

$\cos 0 = 1$

$\therefore C + 3 = 5$

$\therefore C = 2$

$\therefore v = 3 \cos 3x + 2$

$x = \int v \,\,dt$

$= \int \left(3 \cos 3x + 2\right)\,\, dt$

$= 3 \cdot \frac{1}{3} \sin 3x + 2t + C$

$= \sin 3x + 2t + C$

$t = 0, x = -3$

ie

$\sin\left(3\cdot 0\right) + 2 \cdot 0 + C = -3$

$\sin 0 = 0$

$\therefore C = -3$

$\therefore x = \sin 3x + 2t - 3$

when $t = \pi$ ,

$x = \sin 3\pi + 2\pi - 3$

$= 0 + 2\pi - 3$

$= \left(2\pi - 3\right) cm$

kurt.physics · Jul 1, 2009

Makro said:
So I didn't realise I didn't do this exercise and I left it for a week and it seems I've forgotten how to do half of it. Here is a scanned copy of the textbook page
I tried and failed on:
Q8, 9, 10, 12, 13

Did the rest, so yeah, help would be much appreciated.

13)

$\dot{x} = \frac{t}{t^2 + 3}$

$t = 0, x = 0$

$x = \int v \,\, dt$

$\therefore x = \int \frac{t}{t^2 + 3} \,\, dt$

$= \int \frac{1}{2} \cdot \frac{2t}{t^2 + 3} \,\,dt$

$= \frac{1}{2} \int \frac{2t}{t^2 + 3} \,\,dt$

$\int \frac{f'(x)}{f(x)} = \log_{e}f(x) + C$

Therefore,

$x = \frac{1}{2} \log_{e}\left( t^2 + 3\right) + C$

$t = 0, x = 0$

ie

$\frac{1}{2} \log_{e}\left( 0^2 + 3\right) + C = 0$

$\therefore C = - \frac{1}{2} \log_{e}\left(3\right)$

$\therefore x = \frac{1}{2} \log_{e}\left( t^2 + 3\right) - \frac{1}{2} \log_{e}\left(3\right)$

when $t = 10$ ,

$x = \frac{1}{2} \log_{e}\left( 10^2 + 3\right) - \frac{1}{2} \log_{e}\left(3\right)$

$\approx 1.77 \,\,m$

Makro · Jul 1, 2009

$x = \int v \,\, dt$

$\therefore x = \int \left(e^t - 1\right) \,\, dt$

$= e^t - t + C$

Everything is clear as day, except for this bit, I thought when you integrate an exponential you use the rule: 1/a e^(ax+b)? Why is integrating e^t just e^t? Thank you for everything else

tommykins · Jul 1, 2009

Makro said:
Everything is clear as day, except for this bit, I thought when you integrate an exponential you use the rule: 1/a e^(ax+b)? Why is integrating e^t just e^t? Thank you for everything else

differentiating e^t gives e^t

sidereal · Jul 1, 2009

Makro said:
Everything is clear as day, except for this bit, I thought when you integrate an exponential you use the rule: 1/a e^(ax+b)? Why is integrating e^t just e^t? Thank you for everything else

it's because in these motion problems we are integrating with respect to t rather than with respect to x

you're right, $\int e^{ax+b}\,\text{d}x = \frac{1}{a}e^{ax+b} + c$

but here we are doing $\int e^{at+b}\,\text{d}t$ and so we treat the t exactly the same way we treated the x in the original rule - i.e. $\int e^t - 1\,\text{d}t = e^t - t + c$

Makro · Jul 2, 2009

Right, I dunno why I didn't see that before, thanks

Motion + Integration (1 Viewer)

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