please help how do i go about solving this??

the_d

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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

use the formula:
[tex]d = V_{average}t[/tex]

the_d

and then I subtract the two to find the final displacement

dextercioby

No,you have to add the 2 distances.

Daniel.

christinono

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

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

Average velocity is total distance divided by total time.

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.

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

christinono

ha ha, very funny

dextercioby

Check your numbers again...

Daniel.

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]

the_d

where did u get the 2 from???

christinono

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]