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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
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 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:
$$d = V_{average}t$$
 and then I subtract the two to find the final displacement

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No,you have to add the 2 distances.

Daniel.

 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.
 thanks dude
 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.

 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.
 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

 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.

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 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 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.
 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

 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

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 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?

Daniel.

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

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

where did u get the 2 from???

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