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finding time given velocity equation and acceleration

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?

Reply 1

Flxmz

It’s a quadratic, so you’re supposed to get two answers

83E25ADB-C979-4194-A6A4-A4B0EB4A02CB.jpg.jpeg11.5KB

Reply 2

kswales1

Original post by Flxmz

It’s a quadratic, so you’re supposed to get two answers

thanks for replying!
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

Reply 3

Sinnoh

Original post by Flxmz

It’s a quadratic, so you’re supposed to get two answers

What you've done there is solve t for when v = 0. That's not what the question is asking.

Original post by kswales1

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.

Reply 4

kswales1

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

Reply 5

Sinnoh

Original post by kswales1

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.

(edited 2 years ago)

Reply 6

Muttley79

Original post by Flxmz

It’s a quadratic, so you’re supposed to get two answers

Please don't post solutions and this isn't correct btw :smile:

Reply 7

kswales1

Original post by Sinnoh

Thanks for your help btw :smile:

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?

Reply 8

Sinnoh

Original post by kswales1

Thanks for your help btw :smile:

I've attached a photo of my working -

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.