How Do You Calculate Total Displacement in Physics?

[the_d]

please help how do i go about solving this??

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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

 

[christinono]

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:
$$d = V_{average}t$$

 

[the_d]

and then I subtract the two to find the final displacement

 

[dextercioby]

No,you have to add the 2 distances.

Daniel.

 

[christinono]

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]

thanks dude

 

[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.

 

[christinono]

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]

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]

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]

Aren't the words "distance travelled" and "displacement" different. I though displacement meant distance and direction from the original starting point.

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]

new question

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]

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]

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]

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),
$$\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}$$

 

[the_d]

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),
$$\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}$$

where did u get the 2 from?

 

[christinono]

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: $$d = V_{ave}t$$

 

[dextercioby]

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

Daniel.

 

[the_d]

your saying I take 30m/s and divide it by 2?? that is not right either.

 

[christinono]

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

Daniel.

I's not "side talking". I'm referring to the "new question". Here, average velocity IS involved in finding the distance.

Anyways, got to go. I hope I didn't irritate you too much Daniel

 

[christinono]

your saying I take 30m/s and divide it by 2?? that is not right either.

You take the 30 m/s, divide it by 2, and multiply it by the time to get the distance.

 

[dextercioby]

Do you know Galileo Galilei's formula for uniformly accelerated motion...??

Daniel.

 

[dextercioby]

Christinono,your method is equivalent to the one i suggested...Let's leave the OP to decide which is better for him to use.
Daniel.

 

[the_d]

galelio

Do you know Galileo Galilei's formula for uniformly accelerated motion...??

Daniel.

yes, those are the four different eq'ns

 

[dextercioby]

yes, those are the four different eq'ns

Nope,it's only one.

$$v_{fin}^{2}=v_{init}^{2}+2ad$$

Daniel.

 

[the_d]

Nope,it's only one.

$$v_{fin}^{2}=v_{init}^{2}+2ad$$

Daniel.[/QUOT
i got it, thanks 4 the help dude