# suvat question

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#1
i have this question that keeps bugging me:
i know that suvat equations apply to questions with constant acceleration, but does this include when a = 0?
0
4 years ago
#2
(Original post by blankboi)
i have this question that keeps bugging me:
i know that suvat equations apply to questions with constant acceleration, but does this include when a = 0?
Yes, the suvat equations work for a = 0, since this is a constant value for a.

When a = 0:

v = u + at --> v = u, which makes sense if there is no acceleration. Same for v^2 = u^2 + 2as.

s = ut + 0.5*at^2 --> s = ut, which is the usual equation for zero acceleration (distance = speed x time).

s = t(u+v)/2 --> s = t(2u)/2 = ut, as above.
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#3
(Original post by ombtom)
Yes, the suvat equations work for a = 0, since this is a constant value for a. All of the equations should reduce to distance = speed x time.
thanks

but this makes me confused after doing this question:
A steel ball of mass 0.15kg, released from rest in a liquid, falls a distance of 0.2m in 5 seconds. Assuming the ball reaches terminal velocity within a fractio of a second, calculate:
i) its terminal speed

I originally used the suvat equations to get speed but i got 0. 08 (answer is 0.04) and im not sure why its wrong since a would be 0.

i know that i need to use the speed = distance / time equation, but this is for constant velocity and that means 0 acceleration.
0
4 years ago
#4
(Original post by blankboi)
thanks

but this makes me confused after doing this question:
A steel ball of mass 0.15kg, released from rest in a liquid, falls a distance of 0.2m in 5 seconds. Assuming the ball reaches terminal velocity within a fractio of a second, calculate:
i) its terminal speed

I originally used the suvat equations to get speed but i got 0. 08 (answer is 0.04) and im not sure why its wrong since a would be 0.

i know that i need to use the speed = distance / time equation, but this is for constant velocity and that means 0 acceleration.
speed = distance/time = 0.2/5 = 0.04 m/s. What were you trying to do? 0
#5
(Original post by ombtom)
speed = distance / time = 0.2/5 = 0.04 m/s. What were you trying to do? I first did s = (u+v)t/2
and so rearranged to (2s/t) = v and this equals to 0.08 which is wrong

when using the speed = distance / time equation, does that mean a = 0?
if it does, then i don't understand why it doesn't work for suvat
0
4 years ago
#6
(Original post by blankboi)
I first did s = (u+v)t/2
and so rearranged to (2s/t) = v and this equals to 0.08 which is wrong

when using the speed = distance / time equation, does that mean a = 0?
if it does, then i don't understand why it doesn't work for suvat
Speed = distance/time only applies when a = 0.

Suvat equations only apply when a = constant (0 is a constant).

Variable acceleration requires differentiation/integration.

s = (u+v)t/2.

As a = 0, u = v, so s = 2ut/2 = ut.

Your mistake was ignoring u; u = v, so u + v = 2u, or 2v.
0
#7
(Original post by ombtom)
Speed = distance/time only applies when a = 0.

Suvat equations only apply when a = constant (0 is a constant).

Variable acceleration requires differentiation/integration.

s = (u+v)t/2.

As a = 0, u = v, so s = 2ut/2 = ut.

Your mistake was ignoring u; u = v, so u + v = 2u, or 2v.
hm i get what your'e saying but it says it was released from rest so u = 0?
0
4 years ago
#8
(Original post by blankboi)
hm i get what your'e saying but it says it was released from rest so u = 0?
It says it accelerates to terminal velocity in a negligible amount of time. After it has reached terminal velocity, a = 0. As it had to accelerate for a short period of time, the answer is an approximation.

We're applying v = d/t from the time when it reaches its maximum velocity.

You're right that the speed starts as 0, but it accelerates quickly; so quickly, that we can consider u to be a positive value.
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