25 Origin 070 070 0.2000 (s) 0.5000 (s) 0.8000 (s) Data for the positi f the of

Question

25 Origin 070 070 0.2000 (s) 0.5000 (s) 0.8000 (s) Data for the positi f the of the h tal bar of the triangl en every tenth of a d d g its fi d of Th rk of fixed meter stick. t (s) x (cm) v cm/s) Triangle Motion 0.000 52 0.100 49.9 0.200 44.5 54.0 50 0.300 39 40 0.400 35.2 0.500 34.8 E 30 0.600 36.9 20 0.700 43.0 0.800 49.2 0.900 53.6 1.000 54.4 0.0 0.2 0.4 0.6 0.8 Examine the po graph of the data she Does the tri ppe y tha the fi d? A Exp Discuss the nature of the motion based on the shape of the graph. At approximately what time, if any, is the triangle changing direction? At approximately what time does it have the greatest The gr ty? Exp for y positi gati Use the data table and the definition o y of the ty to h of the ti 0.100 d 0.900 In th p frog triangle at ethod the posi bef the ind ted d the position just after the indicated ur calculation. For example, to calculate th -0.100 44.5 52 th the diff of the ti nd t Hint: Only times and positions in the yellow boxes to get a velocity in a yellow box and only use times and hite boxes to g hite box. A ulati of th positi 0.100 by 44.5 38.0 0.200 s 0.000 s dsheet for y nd sub f the data b pa y refer to being dista d be peop alcul r/te, xalt, etc. thod f finding the at the diff Try g tha thod of call f ye for yo Oftel oddly shaped b bly th graph btained fr data fifth order poly al that fits this data p poly t 803 38.5 472 376 x 9.25 52 Triangle Motion 30 20 0.0 0.2 0.4 0.6 0.8 -thry Using this polynomial approximation, find the instantaneous velocity at t 0.600 s. To do this properly yo ed to tak bstituti ppropriate fully. (If you ha d h of a poly al y alcul Secti Unit 3 of the Activity Guid 0.600 Are th ty y pect? (Finding the difference or discrepancy, as described the C Activity 23 the A ty Guide good the two y to value

Explanation / Answer

There are multiple questions here. So my intention will be to help you understand the graph and in this way we will be answering 4 of the questions.

a)

x - t graph is parabola facing up.

Slope of the curve gives velocity. So Initially slope is negative and it starts decreading in magnitude untill 0.5 s after which slope becomes positive and starts increasing.

So, velocity was negative first and it decreases till 0.5 s and after that it becomes positive and starts increasing.

so, velocity is not constant.

But acceleration is constant since slope increases by same amount in each time interval.

b)I already discussed the nature of motion above. It canges direction at time = 0.5 s since velocity changes direction here.

Maximum negative velocity is at begining around 0 s and maximum positive velocity is around 1 s

c)

average velocity = (x2-x1)/(t2-t1)

please calculate it for all time interval. I have already mentioned how this velocty will change.

d)

No.

Velocity is change in distance/ change in time. It is not distance / time. If you are using distance/time,it will give average velocity from t=0 and not for particular block.

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.