Question

The positon of a runner is given by x=5.40t-0.580t^2. Where x is in meters and t is in seconds. What is the average speed between time t=0 and t=9.31s?

Answer 1

May I answer this?

Answer 2

Average speed is distance/time. You can calculate distance by subtracting the initial position from the final position. Also, watch out for the direction change!

Answer 3

I cant find an answer . I am getting 2x10^-4. please help

Answer 4

Hello! Is the equation \(x = 5.4t^2 - 0.58t^2\)?

Answer 5

Yes

Answer 6

OK. So, what is the position at x = 9.31?

Answer 7

you mean t=9.31

Answer 8

you mean at time t - 9.31 seconds?

Answer 9

Whoops.

Answer 10

Yeah. I am still used to that x-y convention. Damn.

Answer 11

the position at time t=9.31s will be 1.782x10^-3

Answer 12

1.862 i mean sorry

Answer 13

I don't know, but that should be right.

Answer 14

But we are asked to find average speed. So what should my next step be??

Answer 15

Uh, why did you get such a small value

Answer 16

I plugged in the value of time t=9.31 seconds in the equation. Isnt that right ?

Answer 17

Doesn't seem so. Are you sure that is. 10^-3?

Answer 18

yes. So i know average speed = total distance traveled/ time taken.
So i plugged in the value for both the times to find the distance traveled but I am getting this. So what should I do according to you?

Answer 19

5.4(9.31)^2 - 0.58(9.31)^2

Answer 20

hey it's just 5.40t not t^2 sorry.

Answer 21

hey u der/??

Answer 22

You weren't quite right. Don't forget the 10. It is 1.862*10^-3. Then you will find the speed for t=0

Answer 23

Hello, I am back. In that case, first be careful: Change in position will not be equal to the distance covered by the object if there is a change in direction.

Answer 24

To find when there are changes in direction if any, find the derivative of the equation and set it =0 and find t

Answer 25

For checking if there is a change in direction, find the derivative of the function and equate it to zero. That'd tell you the time when the object is changing its direction (let us call that \(t_0\). For finding the distance covered, find the displacement upto \(t_0\) and then after \(t_0\).

Answer 26

Erp there we go. I'll leave this to Kohli here. Good luck.

Answer 27

Thank you I will try this and let you know

Answer 28

Hey kohli, I took the derivate and I found time t=4.655 seconds. can you please guide me after this ??

Answer 29

how do I find the displacement upto time t(o) ??? and after t(0)??

Answer 30

He keeps disappearing... Well basically this means that the runner will run one direction from t=0 to t=4.655 and then go back until t=9.31.

Answer 31

oh ok thanks .. so basically I should Find distance for t-0 to t=4.655,? and then go back until t=9.31s ? what does that mean /? can you please elaborate. Also what should the total time travelled be then/

Answer 32

Yup! Distance is the path covered, rather than the position. Now, if I go two steps forward, then go two steps back, my position change is zero. But I have stepped four steps nevertheless. So how am I supposed to find the distance covered? Apparently, the time when the thing changes its direction is very crucial. So, until I changed my direction, I stepped two times. After changing my direction, I stepped twice. That would amount to 2 + 2 = 4.

Answer 33

Similarly, you have to find the distance it covered in the time before it changed it direction. After changing its direction, find the distance it covered after that.

Answer 34

Yeah I understand when you explain me with steps. But then for this particular example. after calculating distance from t=0 to t=4.65 what Should I do next? sorry I am a bit confused here

Answer 35

You have calculated the distance from t = 0 to t = 4.65 Now, find it from t = 4.65 to t = 9.31, as you have to find it from t = 0 to 9.31.

Answer 36

Ok and then should i divide it by 9.31 seconds? as that's the total time for the journey

Answer 37

Yes, that's correct.

Answer 38

So, to conclude, the total distance from 0 to 9.31 seconds is the displacement from 0 to 4.65, plus the displacement from 4.65 to 9.31. The total time is 9.31, so divide the total distance by that to get the average speed.

Answer 39

It's giving me worng. My asnwer is 1.35 x 10^5

Answer 40

Let me see.

Answer 41

12.5 for 0 to 4.65 and 12.5 for 4.65 to 9.31

Answer 42

25 in totality. Divide that by the time.

Answer 43

oh ok i made some calculation error . Thanks Parth. Sorry for the trouble. But i really appreciate your help.

Answer 44

Haha, no sweat! I enjoyed this question.