# finding time given velocity equation and acceleration

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#1
I seem to be posting on this forum a lot lol
Anyway, i need help regarding the following question:

I'm trying to do part A right now, and what's throwing me off is how it says "valueS" of t (emphasis on the plural!) and how it's worth 4 marks which makes me think I've done something wrong as my working seemed too easy.

Here's my working:
acceleration = dv/dt, so a = -2t + 2.
The question tells me that a = 1, so I put that into a = -2t + 2 to solve for t, which gets me t = 1/2.
Have I done something wrong here or am I just overthinking?
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1 month ago
#2
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#3
(Original post by Flxmz)
I'm just a bit confused by your answer though - I don't see where the acceleration is taken into account? The question seems to imply that I'd have to use acceleration to find t somehow
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1 month ago
#4
(Original post by Flxmz)
What you've done there is solve t for when v = 0. That's not what the question is asking.

(Original post by kswales1)
I'm just a bit confused by your answer though - I don't see where the acceleration is taken into account? The question seems to imply that I'd have to use acceleration to find t somehow
The extra value is hidden in the word "magnitude" - if the magnitude of the acceleration is 1, then the acceptable values of acceleration are 1 and -1.
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#5
(Original post by Sinnoh)
What you've done there is solve t for when v = 0. That's not what the question is asking.

The extra value is hidden in the word "magnitude" - if the magnitude of the acceleration is 1, then the acceptable values of acceleration are 1 and -1.
Ah that makes sense. So I'd calculate t for when a = 1 AND when a = -1.
Do you mind giving me some guidance for part b as well? I've attempted it, but I'm stuck again.
I assume to find the distance (s), I'd have to use s = ∫ v dt, which gives me s = 8t + t^2 - t^3 /3 + C. However, this means I have to find the integration constant (C) and I'm not sure how to do that
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1 month ago
#6
(Original post by kswales1)
Ah that makes sense. So I'd calculate t for when a = 1 AND when a = -1.
Do you mind giving me some guidance for part b as well? I've attempted it, but I'm stuck again.
I assume to find the distance (s), I'd have to use s = ∫ v dt, which gives me s = 8t + t^2 - t^3 /3 + C. However, this means I have to find the integration constant (C) and I'm not sure how to do that
Well, at t = 0 the distance from O should be 0, right? So you use your boundary conditions to find that constant of integration.

Make a sketch of displacement against time; remember that distance and displacement mean different things and you may need to calculate them differently.
Last edited by Sinnoh; 1 month ago
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1 month ago
#7
(Original post by Flxmz)
Please don't post solutions and this isn't correct btw
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#8
(Original post by Sinnoh)
Well, at t = 0 the distance from O should be 0, right? So you use your boundary conditions to find that constant of integration.

Make a sketch of displacement against time; remember that distance and displacement mean different things and you may need to calculate them differently.
I've attached a photo of my working -
I'm not sure if i've made a mistake somewhere because from my working, at t = 4, s = 26.7 however at t = 5, s = 23.3, which doesn't seem to make sense. I'm also confused because aren't parts B and D asking for the same thing?
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1 month ago
#9
(Original post by kswales1)
I've attached a photo of my working -
I'm not sure if i've made a mistake somewhere because from my working, at t = 4, s = 26.7 however at t = 5, s = 23.3, which doesn't seem to make sense. I'm also confused because aren't parts B and D asking for the same thing?
Remember that the velocity becomes negative after t = 4, so from then on it's moving backwards. That's why the displacement has gone down. To calculate distance you need to make any negative displacement positive.

The difference between distance and displacement is that if you walk 5m to the right and then back 5m to the left, your displacement is 0 but your distance travelled is 10m.
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#10
(Original post by Sinnoh)
Remember that the velocity becomes negative after t = 4, so from then on it's moving backwards. That's why the displacement has gone down. To calculate distance you need to make any negative displacement positive.

The difference between distance and displacement is that if you walk 5m to the right and then back 5m to the left, your displacement is 0 but your distance travelled is 10m.
So would this be correct for part D?

26.7 (distance at t=4) - 23.3 (distance at t=5) = 3.4, which means the total distance is 26.7 + 3.4 = 30.1?
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1 month ago
#11
(Original post by kswales1)
So would this be correct for part D?

26.7 (distance at t=4) - 23.3 (distance at t=5) = 3.4, which means the total distance is 26.7 + 3.4 = 30.1?
Yeah, but be careful with your significant figures, keep it exact until the very end - the distance is exactly 30, not 30.1.
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#12
(Original post by Sinnoh)
Yeah, but be careful with your significant figures, keep it exact until the very end - the distance is exactly 30, not 30.1.
Right, thanks so much for your help!!
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