can someone explain this working out for finding distance travelled. (1 Viewer)

ymcaec

Member
The area under a velocity-time curve is the displacement, which in many cases, is the distance travelled.

girlworld_club

Member
i was wondering where they got 2t2 in the bracket while integrating.

braintic

Well-Known Member
i was wondering where they got 2t2 in the bracket while integrating.
It could well have been 2t^2 - 6 (the displacement function). But after substituting and subtracting, the 6s cancel. Which is the reason you don't need a constant in the definite integral.

There wasn't actually any integration done, so it was a rather dumb way of setting it out. It would have been easier to say:
distance = x(5) - x(0) = 44-(-6) = 50

The note in the margin is actually rubbish. Method 2 would give exactly the same answer if the particle is oscillating, because the calculation is essentially the same. In this case, you would have to break the motion into separate domains with +ve and -ve velocities, and this would be a necessary and sufficient requirement for either method. I hope your teacher gave you an example of this.

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AnimeX

Member
i was wondering where they got 2t2 in the bracket while integrating.
x=2t^2 -6
v=x' = 4t
when you integrate velocity with respect to time ie v(t) you're integrating 4t, which is why there's a 2t^2

D94

Well-Known Member
I think they differentiated the displacement-time equation, which equals the velocity-time equation = 4t. Then they integrated that which = 2t^2, evaluated from t = 0 to t = 5.

Edit: too slow.

girlworld_club

Member
didn't you say distance was area under velocity curve? how could have they have used the displacement function.

girlworld_club

Member
I have another question on distance. if anyone could help me.

what is the total distance travelled by x = 2sint-t from pi to zero. (using the second method).
and total distance travelled by the velocity curve 2-t from t1 to 2, and from 2 to t2 (answer is in terms of t1 and t2).
The the second question my only problem is setting out and not so much integrating and plugging in values.

anyone care to explain??

ymcaec

Member
I have another question on distance. if anyone could help me.

what is the total distance travelled by x = 2sint-t from pi to zero. (using the second method).
and total distance travelled by the velocity curve 2-t from t1 to 2, and from 2 to t2 (answer is in terms of t1 and t2).
The the second question my only problem is setting out and not so much integrating and plugging in values.

anyone care to explain??
$\bg_white Distance travelled by x = 2\sint-t from 0 - \pi seconds.\\1. Firstly, find the velocity equation.\\x = 2 \sin t - t\\\dot{x}=2 \cos t - 1\\\\Because the velocity graph crosses the x-axis between 0 and \pi, and we want \textit{distance travelled} not \textit{displacement}, we have to consider the curve as two parts.\\\\First find where the graph cuts the x-axis (when \dot{x} = 0)\\\\2 \cos t - 1=0\\\cos t = \frac{1}{2} \\ \therefore t = \frac{\pi}{3} (0

$\bg_white \int_{0}^{\frac{\pi}{3}} (2 \cos t - 1) dt + \left | \int_{\frac{\pi}{3}}^{\pi} (2 \cos t - 1) dt \right |\\= \left [ 2\sin t-t \right ]_{0}^{\frac{\pi}{3}}+ \left | \left [ 2\sin t-t \right ]_{\frac{\pi}{3}}^{\pi} \right |\\=\left ( 2\sin \frac{\pi}{3} - \frac{\pi}{3} \right ) - \left ( 2\sin 0 - 0 \right ) + \left | \left ( 2\sin \pi - \pi \right ) - \left ( 2\sin \frac{\pi}{3} - \frac{\pi}{3} \right ) \right |\\=\left ( 2\sin \frac{\pi}{3} - \frac{\pi}{3} \right ) + \left | - 2\sin \frac{\pi}{3} - \frac{2\pi}{3} \right |\\= 2\sin \frac{\pi}{3} - \frac{\pi}{3} + 2\sin \frac{\pi}{3} + \frac{2\pi}{3}\\= 4\sin \frac{\pi}{3} + \frac{\pi}{3}\\= 2\sqrt{3} + \frac{\pi}{3}$

and the second one is similar - split it into two parts

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