# Suvat equations

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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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#2

(Original post by

When using s=ut+1/2at^2 to find t, but both values are positive, which value do I use for t?

**Intramoon**)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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(Original post by

depends on the context of the q - can we see the q?

**HS_1**)depends on the context of the q - can we see the q?

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#4

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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?”

**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?”

Up to you to decide which one of your solutions is more appropriate.

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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.

**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.

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#6

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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?

**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?

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Can you think of another suvat equation which involves u, v, a and s to solve for u? Then determine the time?

**mqb2766**)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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#8

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I can think of v^2=u^2+2as but not quite sure how to work out v and u with that 😅

**Intramoon**)I can think of v^2=u^2+2as but not quite sure how to work out v and u with that 😅

v^2 - u^2 = 2as

Substitute v then solve for u.

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(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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#10

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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

**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

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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(Original post by

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.

**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.

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#12

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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..

**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..

A simple sketch would be enough to understand.

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(Original post by

You could investigate? Extra solutions mean something.

A simple sketch would be enough to understand.

**mqb2766**)You could investigate? Extra solutions mean something.

A simple sketch would be enough to understand.

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#14

(Original post by

Is it the time for the particle going through the same situation but before the start of the question?

**Intramoon**)Is it the time for the particle going through the same situation but before the start of the question?

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