- #1
JanineSamson
- 3
- 0
Hi Everyone,
I haven't had any Physics education so I'll apologise in advance because this question is probably going to be stupid.
If I had something starting from rest (u = 0) and I have a distance (s = 0.8m), and time it takes to travel that distance (t = 1.9s), which way would you go about working out the acceleration?
I can see two ways, but one of them is obviously incorrect because you end up with different answers.
1.
Use
[tex]s = ut + 1/2at^2[/tex]
Since u = 0
[tex]s = 1/2at^2[/tex]
rearranging...
[tex]a = 2s/t^2[/tex]
Which with my numbers would give, a = 1.6/3.61 = 0.443ms^-22.
Find the velocity using s/t, so v = 0.8/1.9 = 0.42ms^-1
then a = (v - u) / t = (0.42 - 0) / 1.9 = 0.221ms^-2
----------------------------------
Is one of those methods correct? I thought they would both come out the same... but they don't so I seem to be doing something horribly wrong.
Any help would be greatly appreciated, many thanks,
Janine.
I haven't had any Physics education so I'll apologise in advance because this question is probably going to be stupid.
If I had something starting from rest (u = 0) and I have a distance (s = 0.8m), and time it takes to travel that distance (t = 1.9s), which way would you go about working out the acceleration?
I can see two ways, but one of them is obviously incorrect because you end up with different answers.
1.
Use
[tex]s = ut + 1/2at^2[/tex]
Since u = 0
[tex]s = 1/2at^2[/tex]
rearranging...
[tex]a = 2s/t^2[/tex]
Which with my numbers would give, a = 1.6/3.61 = 0.443ms^-22.
Find the velocity using s/t, so v = 0.8/1.9 = 0.42ms^-1
then a = (v - u) / t = (0.42 - 0) / 1.9 = 0.221ms^-2
----------------------------------
Is one of those methods correct? I thought they would both come out the same... but they don't so I seem to be doing something horribly wrong.
Any help would be greatly appreciated, many thanks,
Janine.
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