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The position of a squirrel running in a park is given by r⃗ =[(0.280m/s)t+(0.0360m/s2)t2]i^+ (0.0190m/s3)t3j^.

At 5.01 s , how far is the squirrel from its initial position?

At 5.01 s , what is the magnitude of the squirrel's velocity?

At 5.01 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

Express your answer to three significant figures and include the appropriate units.

 Sep 8, 2015
edited by Guest  Sep 8, 2015

Best Answer 

 #1
avatar+26396 
+20

The position of a squirrel running in a park is given by r⃗   = [(0.280m/s)t+(0.0360m/s2)t2]i+ [ (0.0190m/s3)t3 ] j

a) At 5.01 s , how far is the squirrel from its initial position?

b) At 5.01 s , what is the magnitude of the squirrel's velocity?

c) At 5.01 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

Express your answer to three significant figures and include the appropriate units.

 

r=[0.280 mst+0.0360 ms2t2]i+[0.0190 ms3t3]jr=(x(t)y(t))=(0.280 mst+0.0360 ms2t20.0190 ms3t3)a) t=5.01 sx(5.01 s)=0.280 ms(5.01 s)+0.0360 ms2(5.01 s)2=2.30640360000 my(5.01 s)=0.0190 ms3(5.01 s)3=2.38927851900 mr(5.01 s)=2.306403600002 m2+2.389278519002 m2=3.32086576173 m

 

The squirrel is 3.32 m from its initial position(0,0)

 

v=(vxvy)=(d x(t)dtd y(t)dt)=(0.280 ms+20.0360 ms2t30.0190 ms3t2)b) t=5.01 svx(5.01 s)=0.280 ms+20.0360 ms2(5.01 s)=0.64072 msvy(5.01 s)=30.0190 ms3(5.01 s)2=1.4307057 msv(5.01 s)=0.640722 (ms)2+1.43070572 (ms)2=1.56762269645 ms

 

The magnitude of the squirrel's velocity is 1.57 m/s

 

c) direction =arctan(vyvx)=arctan(1.4307057 ms0.64072 ms)=arctan(2.23296556998)=65.8754973481

 

The direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity is 65.9 degrees

 

laugh

 Sep 8, 2015
 #1
avatar+26396 
+20
Best Answer

The position of a squirrel running in a park is given by r⃗   = [(0.280m/s)t+(0.0360m/s2)t2]i+ [ (0.0190m/s3)t3 ] j

a) At 5.01 s , how far is the squirrel from its initial position?

b) At 5.01 s , what is the magnitude of the squirrel's velocity?

c) At 5.01 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

Express your answer to three significant figures and include the appropriate units.

 

r=[0.280 mst+0.0360 ms2t2]i+[0.0190 ms3t3]jr=(x(t)y(t))=(0.280 mst+0.0360 ms2t20.0190 ms3t3)a) t=5.01 sx(5.01 s)=0.280 ms(5.01 s)+0.0360 ms2(5.01 s)2=2.30640360000 my(5.01 s)=0.0190 ms3(5.01 s)3=2.38927851900 mr(5.01 s)=2.306403600002 m2+2.389278519002 m2=3.32086576173 m

 

The squirrel is 3.32 m from its initial position(0,0)

 

v=(vxvy)=(d x(t)dtd y(t)dt)=(0.280 ms+20.0360 ms2t30.0190 ms3t2)b) t=5.01 svx(5.01 s)=0.280 ms+20.0360 ms2(5.01 s)=0.64072 msvy(5.01 s)=30.0190 ms3(5.01 s)2=1.4307057 msv(5.01 s)=0.640722 (ms)2+1.43070572 (ms)2=1.56762269645 ms

 

The magnitude of the squirrel's velocity is 1.57 m/s

 

c) direction =arctan(vyvx)=arctan(1.4307057 ms0.64072 ms)=arctan(2.23296556998)=65.8754973481

 

The direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity is 65.9 degrees

 

laugh

heureka Sep 8, 2015

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