Question

A particle moving in a straight line with constant retardation of 3ms^-2 has an initial velocity of 10ms^-1. find after what time it returns to its starting point

Answer 1

@Hero @dpaInc

Answer 2

?

Answer 3

@SmoothMath @phi @radar

Answer 4

time is 10/3?

Answer 5

i dont understand ><

Answer 6

can u solve it using
s= 1/2at^2 + ut?

Answer 7

what is S0?

Answer 8

we dont know initial postition?

Answer 9

so S = S0

Answer 10

@Callisto

Answer 11

returns to its starting point implies that displacement = 0 ?!

Answer 12

yes
0 = ut+ 1/2at^2

solve for t is it?

Answer 13

I'm thinking in this way. Not sure if it is correct though. I did bad in Physics :(

Answer 14

i got it :) thanks

Answer 15

welcome :)

Answer 16

we could think like this: the particle goes right at v= 10 m/s , but slowing by -3 m/s^2
the time it takes to slow to zero is given by
\[ \Delta v= a t\]
or 10= 3t, t= 10/3 seconds
we could argue symmetry to say the return will also take 10/3 seconds for a total time of 30/3 or 6 2/3 seconds

or we could say, after 10/3 of a second the particle has gone
s= 0.5a t^2 + vt meters
s= 0.5(-3)*100/9 + 10*10/3 = 50/3 meters
the return trip is the same distance, with the initial velocity zero:
50/3 = 0.5(-3)* t^2 --> t=10/3

or we could graph the problem,