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Carnegie Mellon University

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Problem 4 Easy Difficulty

An ant crawls on the floor along the curved path shown in Figure $P 3.4$ . The ant's positions and velocities are indicated for times $t_{i}=0$ and $t_{f}=$ 5.00 s. Determine the $x-$ and $y$ -components of the ant's (a) displacement, (b) average velocity, and (c) average acceleration between the two times.

Answer

(a) $( \Delta x , \Delta y ) = ( 0.3,0.2 )$
(b) $\Delta \vec { v } _ { n v , x } = 0.06 \mathrm { m } 8 ^ { - 1 }$ and $\Delta \vec { v } _ { m v } = 0.04 \mathrm { m } \mathrm { s } ^ { - 1 }$
(c) $a _ { \mathrm { av } , \mathrm { x } } = 0.02 \mathrm { ms } ^ { - 2 }$ and $a _ { \mathrm { avy } } = 0.02 \mathrm { ms } ^ { - 2 }$

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Kervin P.

February 18, 2021

A miniature quadcopter is located at xi = 2.00 m and yi = 4.50 m at t = 0 and moves with an average velocity having com- ponents vav,x = 1.50 m/s and vav,y = 21.00 m/s. What are the (a) x - coordinate and (b) y - coordinate of the quadcopter’s position

Video Transcript

so the horse for party, the horizontal component of the displacement would be X final minus X initial. This would be equal in point for 00 minus 00.100 This was giving us 0.3 serial zero meters delta. Why, this is B equaling my final minus y initial. This is equaling 0.350 meters minus 0.150 meters. This is equal in 0.200 meters. Disappear to answers for party for part B, the average velocity in the X direction would be Delta X divided by T For Dr T. This is equaling 0.300 meters divided by five seconds. And so this is equaling 50.60 meters per second and for the the average velocity in the wider actually this would be delta y divided by Delta T, and this is equaling 0.200 meters divided by again five seconds and this is equaling point 040 meters per second. These have you two answers for part beat. Ah, we want to find that the average acceleration we can say that the average acceleration in the extraction this would be equaling 0.1 00 meters per second, divided by 5.0 seconds. This is equaling 0.2 meters per second squared and then for the mother average acceleration in the wider action, this would be equaling negative 0.100 meters per second, divided by 5.0 seconds. And this is equaling negative 0.0 two meters per second. Squared used to be a two answers for part C. That is the end of the solution. Thank you for watching.

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