1- what\'s is the unicyclist\'s total displacement ? explain please (a)1.35 m (b

Question

1- what's is the unicyclist's total displacement ? explain please

(a)1.35 m (b) -1.35 m (c) -2.70 m (d) 2.70 m (e) 0.00 m


2- whar is the total distance that the unicyclist traveld ? explain please

(a)1.35 m (b) -1.35 m (c) -2.70 m (d) 2.70 m (e) 0.00 m

3. what is the value of the avarge velocity ? present your reasoning

(a)-1.8 m/s (b) -0.9 m/s (c) 0.0 m/s (d) 0.9 m/s (e) 1.8 m/s

4- what is the value of the avarge speed? present your reasoning
(a)-1.8 m/s (b) -0.9 m/s (c) 0.0 m/s (d) 0.9 m/s (e) 1.8 m/s

5. what is his acceleation from t= 1.0 s to t=2.0s?

(a)-1.2 m/s^2 (b) -0.6 m/s^2 (c) 0.0 m/s^2 (d) 0.6 m/s^2 (e) 1.2 m/s^2

PART 2. In this part of the assignment, consider a u nicyclist riding along a sidewalk. When he comes to back up immediately and retrace his motion backward. He completes s. His position and velocity at half-second intervals are given by the g table. Note that SI units are used for all quantities. Also note that to a bump he decides this maneuver in 3 second data shown in the following table. Note that SI units are use he is initially moving in the-x direction. Time 1 (s) Time t (s) Position x (m) 1.35 0.60 0.15 0.00 0.15 0.60 1,35 Velocity v (m/s) 0.0 0.5 1.0 1.5 2.0 2.5 1.2 0.6 0.0 0.6 1.2 1.8 3.0 1.35

Explanation / Answer

1)Average velocity is total DISPLACEMENT divided by time.
His displacement is zero he is still at his starting point so the average velocity is zero.
2)
Speed is distance over time
He travelled forwards 1.35 m then back 1.35 m so the total distance is 2.7 m

3)Average velocity is total DISPLACEMENT divided by time.
Since total displacement is 0 hence it is 0

4)AverageSpeed is 2.7m/3,0 s= 0.9 m/s

5)The acceleration is the slope of the velocity time graph.
As it is a straight line the acceleration is constant at all times.

The value is -3.6 m/s / 3.0 sec
= -1.2 m/s^2

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