Chapter 04, Problem 012 At one instant a bicyclist is 31.0 m due east of a park\

Question

Chapter 04, Problem 012 At one instant a bicyclist is 31.0 m due east of a park's flagpole, going due south with a speed of 11.0 m/s. Then 23.0 s later, the cyclist is 31.0 m due north of the flagpole, going due east with a speed of 11.0 m/s. For the cydist in this 23.0 s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration? (Give all directions as positive angles relative to due east, where positive is meaured going counterdockwise.) (a) Number (b) Number (c) Numbeir (d) Number (e) Number (f) Number Click if you would like to Show Work for this question: Units Units Units Units Units Open Show Work

Explanation / Answer

First of all draw a neat and nice cartesian grid and put the flagpole at the origin.

Now the problem says, "at one instant a bicyclist is 31.0 m due east of a park's flagpole" says that at time T=0, the bicyclist's position was (31.0, 0)

"going due south with a speed of 11.0 m/s" says that at time T=0, the bicyclist's velocity was <0, -11.0>


"23.0 seconds later, the cyclist is 31.0 m due north of the flagpole" says that at time T= 23.0, the cyclist is now at (0, 31.0)

"going east with a speed of 11.0 m/s" says that at time T=23, the bicyclist's velocity is <11.0, 0>

So, the displacement is the vector from the starting point of (31.0, 0) to the ending point of (0, 31.0). That vector is <0-31, 31-0> which simplifies to <-31, 31>.

(a) Magnitude of <-31, 31> is (-31)^2+ 31^2) = 43.83 meter.

(b) the direction of <-31, 31> is northwest, which would be 135 degrees (0 is east, 90 is north, 180 is west. 135 is halfway between 90 and 180)

(c) To get the average velocity over the 23 seconds, we can actually ignore the starting and ending velocities.
The average velocity is just the displacement vector, divided by time, so it is <-31/23, 31/23>

(c) Magnitude of <-31/23, 31/23> = 1.90 m/s

(d) Direction of <-31/23, 31/23> is the same as the direction of <-31, 31> was: northwest, at 135 degrees.

(e) For the average acceleration, we have to take the ending velocity, minus the starting velocity, and divide all that by time. (<11 0> - <0 -11> ) / 23 gives us <11 11> / 23, which is <11/23 11/23>.

(e) Magnitude of <11/23 11/23> = (11/23*11/23 + 11/23*11/23) = 0.68 m/s^2

(f) The direction of <11/23 11/23> is northeast, at 45 degrees.

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