You're looking for the period of time that the particle is further than 8m away from it's release. So 8m is a crucial distance and cant' be replaced with some greater distance.

It's important you've got a full appreciation of what's happening to the particle... Have you draw a picture of the problem? Could you describe to me what will happen to it (omitting numbers)? [If you're able to describe the particle travel without numbers accurately, then it makes breaking down and solving these problems easier.]

Reply 26

GabiAbi84

21

Original post by Risermax

Btw if it's more than 8 metres away can I just write any value above 8 like 9 or 20 metres

I feel like you might not be understanding the question itself.

Imagin you’re standing at the bottom of a slide with a tennis ball. You roll the ball up the slide. The ball decelerates as it climbs the slide. At a certain point, the speed will be zero and it will stop and roll back down the slide towards you again.
What the question is asking is for the time that the ball (particle) is further away than 8 metres from the starting point. So it passes the 8 m on the way up, hits the top of its track and comes back down, again passing the 8m mark.

(edited 3 years ago)

Reply 27

DFranklin

18

Original post by Risermax

I did s = ut+1/2at^2
Subbed in the values to get
8 = 12t+2.5t^2
Then did 2.5t^2+12t-8 = 0
Did the quadratic formula and got 0.5933...
So not 0.44 my bad

What do you mean by "did the quadratic formula"?

There are 2 roots to the quadratic. The ball will be 8m away at 2 points in time. If you think about the time you are trying to find, you should be able to see that you need *both* roots (and then to do a calculation involving both of them).

Edit: (and your acceleration has the wrong sign as previously mentioned).

(edited 3 years ago)

Reply 28

Risermax

OP

8

Original post by GabiAbi84

I feel like you might not be understanding the question itself.

Imagin you’re standing at the bottom of a slide with a tennis ball. You roll the ball up the slide. The ball decelerates as it climbs the slide. At a certain point, the speed will be zero and it will stop and roll back down the slide towards you again.
What the question is asking is for the time that the ball (particle) is further away than 8 metres from the starting point. So it passes the 8 m on the way up, hits the top of its track and comes back down, again passing the 8m mark.

No worries I did the question myself and got the answer right anyways it was 3.2 seconds 🙂

Reply 29

LukeEllis2312

1

do you have the answer now mate?

Reply 30

labsose

1

Don't you need to find the difference between the time it takes to come to rest from the gravitational deceleration and that value, that's what I did?My final answer I got was 0.631s.

Reply 31

jelam_2

1

What is the calculation after you have got the two roots?Cuz I got 0.8 and 4 as my roots

Reply 32

_nn.o_

10

Original post by Risermax

Please help with this question: A particle is projected at a speed of 12ms-1 up a straight inclined track. Whilst on the track the particle experiences a constant acceleration down the track of 5ms-2. Find the length of time that the particle is more than 8 metres from the point of projection.

I worked it out using s=ut+1/2at^2 and I got t=4 or 0.8

Reply 33

MatrixOfficial

1

Its not that difficult. Use the formula s=ut+1/2 at^2 where u=12m/s, a=-5m/s^2 (as the acceleration is 5m/s going down and the particle is up), s=8s. That should give you the quadratic equation -2.5t^2+12t+8. If you solve this, you get t=4s and t=0.8s. So during these times, the displacement=8m. So the next step I believe is to find where the displacement is greater than 8m by writing an inequality which is 0.8<t<4 as if you substitute values of t which lie in this range into the s=ut+1/2at^2, the displacement is greater than 8m.

Reply 34

Muttley79

20

Original post by MatrixOfficial

Its not that difficult. Use the formula s=ut+1/2 at^2 where u=12m/s, a=-5m/s^2 (as the acceleration is 5m/s going down and the particle is up), s=8s. That should give you the quadratic equation -2.5t^2+12t+8. If you solve this, you get t=4s and t=0.8s. So during these times, the displacement=8m. So the next step I believe is to find where the displacement is greater than 8m by writing an inequality which is 0.8<t<4 as if you substitute values of t which lie in this range into the s=ut+1/2at^2, the displacement is greater than 8m.

As this was posted three years ago I'm sure the OP has worked it out by now :smile:

Maths: Kinematics

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