Displacement in the 5th second.

[zorro]

Homework Statement

A particle has an initial velocity of 9m/s due east and a constant acceleration of 2 m/s² due west. The distance covered by the particle in the fifth second of its motion is?

The Attempt at a Solution

Using the formula for the displacement in the nth second,
Sn = u + a(n-0.5)

I got S₅=0.

The answer given is 0.5m.

 

[ehild]

The answer given is wrong.

ehild

 

[zorro]

thanks!

 

[Hurkyl]

Actually, the answer in the book is correct.

You didn't answer the question asked -- you gave the answer to a similar but different question.

(hint: zero is obviously a wrong answer to the question asked)

 

[ehild]

Thanks Hurkyl! As it was "displacement" in the title, I did not recognise that "distance covered" was distance travelled, and mixed it with displacement. In this case, the book is right.

ehild

 

[zorro]

Damn! even I thought it is displacement. I got the answer now.
This is how I did it -
observe that the velocity at the beginning of the 5th second is 1m/s and at its end is -1m/s. So we have to find the displacement for half second by symmetry and double it to get the distance travelled.

s = 1/2 -1/2*2*1/4 = 1/4m
Hence total distance is 0.5m. Is there any other method?

 

[ehild]

The distance traveled is equal to the integral of the speed (magnitude of the velocity, |v|) with respect to time. v is positive till t=4.5 s, and negative afterwards, so the integral is split into ∫(9-2t)dt from 4 to 4.5 and ∫(2t-9)dt from 4.5 to 5.

ehild