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Please help how do i go about solving this?by the_d
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#1
Feb605, 10:00 PM

P: 129

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A jogger runs in a straight line with an average velocity of 4:7 m/s for 5.1 min, and then with an average velocity of 4.1 m/s for 3 min. What is her total displacement? Answer in units of meters 


#2
Feb605, 10:01 PM

P: 213

[tex]d = V_{average}t[/tex] 


#3
Feb605, 10:06 PM

P: 129

and then I subtract the two to find the final displacement



#4
Feb605, 10:07 PM

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P: 11,915

Please help how do i go about solving this?
No,you have to add the 2 distances.
Daniel. 


#5
Feb605, 10:09 PM

P: 213




#6
Feb605, 10:10 PM

P: 129

thanks dude



#7
Feb605, 10:11 PM

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Even if the runs in the opposite direction on the second time interval,the TOTAL distance would still be gotten by adding the 2.
Daniel. 


#8
Feb605, 10:13 PM

P: 213




#9
Feb605, 10:14 PM

P: 129

for the same question, to get the avg. velocity during the time what should i do? i know it is not just to add the 2 velocities given and divide by 2



#10
Feb605, 10:15 PM

P: 213




#11
Feb605, 10:20 PM

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Daniel. 


#12
Feb605, 10:21 PM

P: 129

A 500kilogram sports car accelerates uni
formly from rest, reaching a speed of 30 me ters per second in 6 seconds. During the 6 seconds, the car has traveled a distance of how many meters?? i got 180m is that correct? i used : distance = avg. velocity * change in time = (30m.s) * 6 = 180m 


#13
Feb605, 10:21 PM

P: 213




#14
Feb605, 10:26 PM

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Daniel. 


#15
Feb605, 10:31 PM

P: 213

[tex]\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}[/tex] 


#16
Feb605, 10:36 PM

P: 129

where did u get the 2 from??? 


#17
Feb605, 10:40 PM

P: 213




#18
Feb605, 10:41 PM

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Kay,guys,let's leave side talks and refer to the initial problem ...
Daniel. 


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