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Please help how do i go about solving this?

by the_d
Tags: solving
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the_d
#1
Feb6-05, 10:00 PM
P: 129
undefinedundefinedundefined
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
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christinono
#2
Feb6-05, 10:01 PM
P: 213
Quote Quote by the_d
undefinedundefinedundefined
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
use the formula:
[tex]d = V_{average}t[/tex]
the_d
#3
Feb6-05, 10:06 PM
P: 129
and then I subtract the two to find the final displacement

dextercioby
#4
Feb6-05, 10:07 PM
Sci Advisor
HW Helper
P: 11,915
Please help how do i go about solving this?

No,you have to add the 2 distances.

Daniel.
christinono
#5
Feb6-05, 10:09 PM
P: 213
Quote Quote by the_d
and then I subtract the two to find the final displacement
Well, it all depends what the direction is. If he runs in the same direction both times, you have to add the 2 displacements to find the total displacement.
the_d
#6
Feb6-05, 10:10 PM
P: 129
thanks dude
dextercioby
#7
Feb6-05, 10:11 PM
Sci Advisor
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P: 11,915
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.
christinono
#8
Feb6-05, 10:13 PM
P: 213
Quote Quote by dextercioby
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.
Aren't the words "distance travelled" and "displacement" different. I though displacement meant distance and direction from the original starting point.
the_d
#9
Feb6-05, 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
christinono
#10
Feb6-05, 10:15 PM
P: 213
Quote Quote by the_d
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
Average velocity is total distance divided by total time.
dextercioby
#11
Feb6-05, 10:20 PM
Sci Advisor
HW Helper
P: 11,915
Quote Quote by christinono
Aren't the words "distance travelled" and "displacement" different. I though displacement meant distance and direction from the original starting point.
Quote Quote by christinono
Average velocity is total distance divided by total time.
I think that u've given yourself the answer. Besides,since in the problem we're not given the directions (so that we would think it as a vector projection problem),it's natural to assume the simpler version,which is what the teacher would want,right...?

Daniel.
the_d
#12
Feb6-05, 10:21 PM
P: 129
A 500-kilogram 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
christinono
#13
Feb6-05, 10:21 PM
P: 213
Quote Quote by dextercioby
I think that u've given yourself the answer. Besides,since in the problem we're not given the directions (so that we would think it as a vector projection problem),it's natural to assume the simpler version,which is what the teacher would want,right...?

Daniel.
ha ha, very funny
dextercioby
#14
Feb6-05, 10:26 PM
Sci Advisor
HW Helper
P: 11,915
Quote Quote by the_d
A 500-kilogram 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?
Check your numbers again...

Daniel.
christinono
#15
Feb6-05, 10:31 PM
P: 213
Quote Quote by the_d
A 500-kilogram 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
Careful... 30m/s is not the average velocity, but rather the final velocity. Remember, average velocity is total displacement divided by the total time, or in this case (if a is constant),
[tex]\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}[/tex]
the_d
#16
Feb6-05, 10:36 PM
P: 129
Quote Quote by christinono
Careful... 30m/s is not the average velocity, but rather the final velocity. Remember, average velocity is total displacement divided by the total time, or in this case (if a is constant),
[tex]\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}[/tex]

where did u get the 2 from???
christinono
#17
Feb6-05, 10:40 PM
P: 213
Quote Quote by the_d
where did u get the 2 from???
Isn't the average the sum of the 2 velocities divided by 2? In this case, the displacement would be: [tex]d = V_{ave}t[/tex]
dextercioby
#18
Feb6-05, 10:41 PM
Sci Advisor
HW Helper
P: 11,915
Kay,guys,let's leave side talks and refer to the initial problem ...

Daniel.


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