Kinematics

bonkysleuth SG

A particle moves in straight line so that t sec after leaving a fixed point O, its velocity, v ms-1, is given by v = 2t + 4e^-t, find

-its distance from O in the third second

I did this time and again and my answer, 12.8m. the answer given to me is 5.34m. Hmmm, strange yea.

I integrated v, and got s = t^2 + (4e^-t)/-1 + c

the value of c = 4 since when t = 0 , s =0

So subbed in 3 into t and got 12.8+++. Did I go wrong anyway in my workings?

OHSheet SG

yeah. TS.

If they ask for the time in the 3rd second.

Do note:

its distance travelled from t=2sec to t=3sec.

so calculate:

for t=2, displacement x= 7.45m

for t=3 displacement x= 12.79

therefore, distance in the 3rd second is 12.79 - 7.45

which is =5.34

bonkysleuth SG

yeah. TS.

If they ask for the time in the 3rd second.

Do note:

its distance travelled from t=2sec to t=3sec.

so calculate:

for t=2, displacement x= 7.45m

for t=3 displacement x= 12.79

therefore, distance in the 3rd second is 12.79 - 7.45

which is =5.34

Hmm, I still don't think that's right. They are asking for distance from O in the 3rd second, not total distance travelled in the 3rd second alone. Your working is for the distance traveled in the 3rd sec and not from O. (which means it's with reference to position at 2nd sec). So shouldn't I just use displacement?

OHSheet SG

yeah. well... if you read my post..

i underlined " distance in the 3rd second", and showed the working from there. So im just stating the answer if the question is phrased in that way- since you didnt get the ans.

perhaps the question was phrased wrongly??

But what you were saying..

if they ask for distance from O in 3rd sec, just sub in t=3.