A particle, traveling at 6.1 m/s decelerates at 1.4 m/s^2. How long will it take to get back to the start?

Answer 1

First find distance travelled before coming to rest due to retardation :
v^2 - u^2 = 2as
0 - (6.1)^2 = 2*-1.4*s
37.21 = 2.8 s
s = 13.289m

Also, time taken to cover this distance
v = u + at
0 = 6.1 - 1.4t
t = 6.1 / 1.4
t = 4.357 m/s


Now find time taken to go back by this distance to reach starting point.
s = 1/2 at^2
13.289 = 1/2 * 1.4 * t^2
13.289 / 0.7 = t^2
t = 4.357 seconds.

So total time = 4.357 + 4.357
= 8.714 seconds

Answer 2

It will continue going in the original direction for 6.1/1.4 = 4.4 sec. It will take exactly that long to get back.

Ans. 8.8 sec (to 2 sig figs)

Answer 3

friend we need more data to solve this problem, here we dont have any thing about the start the condition...
if it is from rest it should take something like 5secs