Speed after body slows down with negative acceleration.

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SUMMARY

The discussion centers on calculating the final speed of a body that decelerates from an initial speed of 12 m/s with a negative acceleration of 2 m/s² over a distance of 8 meters. The correct approach involves using the equation v² = u² - 2as, leading to a final speed of 10.58 m/s. The forum participants clarify that the quadratic equation derived from the motion equations yields two time values, but only one is physically meaningful in this context. The discussion emphasizes the importance of using the correct kinematic equations for retarding motion.

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Homework Statement


Hi. I know this is a simple problem but i want to know if I am correct. Here it goes.
Body moves with speed 12 m/s. It starts to slow down with negative acceleration 2 m/s[tex]^{}2[/tex] for 8 meters. What is his speed after those 8 meters.

Homework Equations


s=(1/2)*a*t*t
v*v=2as
v=v' + at
s=v'*t + (1/2)*a*t*t

The Attempt at a Solution


s=(1/2)*a*t*t
t*t=2s/a=16/2=8
t=2,82

v=v' - at = 12-5,65=6,34 m/s
that's one solution the other would be.
__________________________________
s=v'*t - (1/2)a*t*t
8=12*t - (1/2)*2*t*t
t*t - 12t +8 = 0
we get quadratic equation... we solve it and get two solutions.
t1=0,71 s
t2=11,29 s

we insert this into
v=v' - at
v1=10,58 m/s
v2=-10,58 m/s

this solution is probably the correct one. But why do we get 2 values for time?
 
Last edited:
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s=(1/2)*a*t*t
t*t=2s/a=16/2=8
t=2,82
This formula is wrong. It should be s = ut + 1/2*at^2.
In the problem t is not given .So you have to use v^2 = u^2 -2as. for retarding motion.
 
rl.bhat said:
s=(1/2)*a*t*t
t*t=2s/a=16/2=8
t=2,82
This formula is wrong. It should be s = ut + 1/2*at^2.
In the problem t is not given .So you have to use v^2 = u^2 -2as. for retarding motion.

Thanx. I understand.
 

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