# Suvat equations

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#1
When using s=ut+1/2at^2 to find t, but both values are positive, which value do I use for t?
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1 month ago
#2
(Original post by Intramoon)
When using s=ut+1/2at^2 to find t, but both values are positive, which value do I use for t?
depends on the context of the q - can we see the q?
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#3
(Original post by HS_1)
depends on the context of the q - can we see the q?
The question was “a particle reduces its speed from 20m/s to 8.2m/s while travelling 100m. Assuming it continues to move with the same constant acceleration, how long will it take to travel another 20m?”
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1 month ago
#4
(Original post by Intramoon)
The question was “a particle reduces its speed from 20m/s to 8.2m/s while travelling 100m. Assuming it continues to move with the same constant acceleration, how long will it take to travel another 20m?”
The particle is constantly reducing its speed and eventually, its speed will become 0. After this point, the particle will travel in the opposite direction. Hence why you have two different positive times at which the particle will have displacement of 20m from its starting point.

Up to you to decide which one of your solutions is more appropriate.
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#5
(Original post by razzor)
The particle is constantly reducing its speed and eventually, its speed will become 0. After this point, the particle will travel in the opposite direction. Hence why you have two different positive times at which the particle will have displacement of 20m from its starting point.

Up to you to decide which one of your solutions is more appropriate.
Ok, thank you! Also, if it’s not too much trouble, can I ask how would I use s=ut+1/2at^2 to solve the question “a particle moves with constant deceleration of 3.6m/s^2 It travels 350m while its speed halves. Find the time it takes to do this” if u is not given? How would you find u?
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1 month ago
#6
(Original post by Intramoon)
Ok, thank you! Also, if it’s not too much trouble, can I ask how would I use s=ut+1/2at^2 to solve the question “a particle moves with constant deceleration of 3.6m/s^2 It travels 350m while its speed halves. Find the time it takes to do this” if u is not given? How would you find u?
Can you think of another suvat equation which involves u, v, a and s to solve for u? Then determine the time?
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#7
(Original post by mqb2766)
Can you think of another suvat equation which involves u, v, a and s to solve for u? Then determine the time?
I can think of v^2=u^2+2as but not quite sure how to work out v and u with that 😅
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1 month ago
#8
(Original post by Intramoon)
I can think of v^2=u^2+2as but not quite sure how to work out v and u with that 😅
So
v^2 - u^2 = 2as
Substitute v then solve for u.
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#9
(Original post by mqb2766)
So
v^2 - u^2 = 2as
Substitute v then solve for u.
Ok, so I did v = 0.5x and u = x
(0.5x)^2 - x^2 = 2x-3.6x350
0.25x^2 - x^2 = -2520
Square root of -2520/0.25-1 = x = 4 square root 210
Substituted that into s = ut+1/2at^2
Two answers were t = 8.050...s or t = 24.152..s which I think is correct, but now I’m not sure which ones more appropriate haha
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1 month ago
#10
(Original post by Intramoon)
Ok, so I did v = 0.5x and u = x
(0.5x)^2 - x^2 = 2x-3.6x350
0.25x^2 - x^2 = -2520
Square root of -2520/0.25-1 = x = 4 square root 210
Substituted that into s = ut+1/2at^2
Two answers were t = 8.050...s or t = 24.152..s which I think is correct, but now I’m not sure which ones more appropriate haha
Easier to use
v = u + at
To get time?

In a sense there could be two solutions as they're talking about speed, rather than velocity. Have a think why.
Last edited by mqb2766; 1 month ago
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#11
(Original post by mqb2766)
Easier to use
v = u + at
To get time?

In a sense there could be two solutions as they're talking about speed, rather than velocity. Have a think why.
Thank you so much I did that and the answer was 8.05s, I don’t know what the t = 24.152s is the time for..
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1 month ago
#12
(Original post by Intramoon)
Thank you so much I did that and the answer was 8.05s, I don’t know what the t = 24.152s is the time for..
You could investigate? Extra solutions mean something.
A simple sketch would be enough to understand.
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#13
(Original post by mqb2766)
You could investigate? Extra solutions mean something.
A simple sketch would be enough to understand.
Is it the time for the particle going through the same situation but before the start of the question?
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1 month ago
#14
(Original post by Intramoon)
Is it the time for the particle going through the same situation but before the start of the question?
Did you sketch it? Not sure how t=24 could be before the start, as t would be negative in that case.
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