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The image with the highlighting and numbers is a solved example problem. The other image is the problem you're solving. Hint:  5m came from the height and then we had 1s. Each square is 5 by 1 or 5m while each triangle is 2.5 m. In other words, rise of 5 and run of 1. On the right side, they become negative because we're going down. One example you add all the 5m and 2.5 m on the left and right side you get 20 m and -20 m. In the example the final answers are 20 m and -20 m and when added together they equal zero. Therefore, the moving man returned to its starting location.   Now solve the Sandy problem the same way. is Sandy correct that the moving man did not return to his starting location? State your claim of agreeing or disagreeing with the moving man having a non zero total displacement. Gather evidence from your calculations. Connect your evidence to your claim by reasoning why it supports your claim. It should be directly linked with your reasoning to why or why not the moving man had a total non-zero displacement.

The image with the highlighting and numbers is a solved example problem. The other image is the problem you're solving. Hint:  5m came from the height and then we had 1s. Each square is 5 by 1 or 5m while each triangle is 2.5 m. In other words, rise of 5 and run of 1. On the right side, they become negative because we're going down. One example you add all the 5m and 2.5 m on the left and right side you get 20 m and -20 m. In the example the final answers are 20 m and -20 m and when added together they equal zero. Therefore, the moving man returned to its starting location.   Now solve the Sandy problem the same way. is Sandy correct that the moving man did not return to his starting location? State your claim of agreeing or disagreeing with the moving man having a non zero total displacement. Gather evidence from your calculations. Connect your evidence to your claim by reasoning why it supports your claim. It should be directly linked with your reasoning to why or why not the moving man had a total non-zero displacement.

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The image with the highlighting and numbers is a solved example problem. The other image is the problem you're solving. Hint:  5m came from the height and then we had 1s. Each square is 5 by 1 or 5m while each triangle is 2.5 m. In other words, rise of 5 and run of 1. On the right side, they become negative because we're going down. One example you add all the 5m and 2.5 m on the left and right side you get 20 m and -20 m. In the example the final answers are 20 m and -20 m and when added together they equal zero. Therefore, the moving man returned to its starting location.

 

Now solve the Sandy problem the same way.


is Sandy correct that the moving man did not return to his starting location?

State your claim of agreeing or disagreeing with the moving man having a non zero total displacement. Gather evidence from your calculations. Connect your evidence to your claim by reasoning why it supports your claim. It should be directly linked with your reasoning to why or why not the moving man had a total non-zero displacement.

Velocity 0 m/s Show Vector 12.0 6.0 -12.0 (0,0) 1 (3,6) (4,6) BER 5 MUN (5.5, 0) 6 (7,-6) 7 MENG TELO MON (8.5, -6) 8 Postion (10,0) Acceleration ++ 10 sec Image shows a Velocity (m/s) Vs Time (s) line graph. The points connecting on the graph are (0,0) (3,6) (4,6) (5.5,0) (7,-6) (8.5,-6) and (10,0).) Using the Moving Man PhET simulator, the above Velocity Vs Time graph was created over a ten second period. A student Sandy analyzed the graph and concluded that the moving man did not return to his original position and had a non-zero total displacement.
Transcribed Image Text:Velocity 0 m/s Show Vector 12.0 6.0 -12.0 (0,0) 1 (3,6) (4,6) BER 5 MUN (5.5, 0) 6 (7,-6) 7 MENG TELO MON (8.5, -6) 8 Postion (10,0) Acceleration ++ 10 sec Image shows a Velocity (m/s) Vs Time (s) line graph. The points connecting on the graph are (0,0) (3,6) (4,6) (5.5,0) (7,-6) (8.5,-6) and (10,0).) Using the Moving Man PhET simulator, the above Velocity Vs Time graph was created over a ten second period. A student Sandy analyzed the graph and concluded that the moving man did not return to his original position and had a non-zero total displacement.
A velocity vs time graph was created using displacement implies that you end at graph imply a total zero displacement? the same spot that you start. Moving = 54 3 Velocity -10 m/s Show Vector X = taf X = 5(²²) X = 2.$ 10.0 Ax = 20m 5.0 2.3 0.0 x = Vit + hat² -5.0 -10.0+ 0 5m/s 5m 5m 5x G= 5 m/s² A=1· w Vi -2 x=vt X=5.3 w. 3 1 5 xi Ax= x₁-x; NOWY WY 2.5-5 5M PA 20 6 AX=-20 m M N -25m- 5mm G=-5~1/²² V ** ty tnhh *** **** ****** ****** ************ ******* R* 9 Acceleration -10 M/S 1440 २६ palma Palmartine 10 sec *** (ww HEELLIS Veda
Transcribed Image Text:A velocity vs time graph was created using displacement implies that you end at graph imply a total zero displacement? the same spot that you start. Moving = 54 3 Velocity -10 m/s Show Vector X = taf X = 5(²²) X = 2.$ 10.0 Ax = 20m 5.0 2.3 0.0 x = Vit + hat² -5.0 -10.0+ 0 5m/s 5m 5m 5x G= 5 m/s² A=1· w Vi -2 x=vt X=5.3 w. 3 1 5 xi Ax= x₁-x; NOWY WY 2.5-5 5M PA 20 6 AX=-20 m M N -25m- 5mm G=-5~1/²² V ** ty tnhh *** **** ****** ****** ************ ******* R* 9 Acceleration -10 M/S 1440 २६ palma Palmartine 10 sec *** (ww HEELLIS Veda
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