Competition - post a photo you've taken of nature >>

M1 - Constant acceleration

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That's what I thought. The train could still keep decelerating to rest after the 6 seconds right?

Yes it looks like it, especially since the first two parts sets you up to find the total time once you eliminate all the variables

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Reply 21

TeeEm

Original post by thefatone

it does but i see no other way than to use time since distance isn't given or mentioned at all in the question.
^^ i take it back
maybe it's worth like a mark or something? or i've don't something horrendously wrong in which case
@Zacken @TeeEm @Student403
i call for your help

sorry I am teaching all day

Reply 22

thefatone

Original post by TeeEm

sorry I am teaching all day

thought so >.>

Reply 23

Zacken

Original post by thefatone

Yep, you've done this one wrong - sorry. :frown:

It's fair simple, actually:

Firs bit of information: s = 30, u = u, v = v, a = a, t = 2

So 30 = 2u + 2a, so 15 = u+a

Second bit of information: s = 30, initial velocity = u + 2a, v = , a = a, t=4

So 30 = 4(u+2a) + 8a

Then solve simul.

Reply 24

thefatone

Original post by Zacken

Yep, you've done this one wrong - sorry. :frown:

all these c2 papers making me compile all my knowledge is kill me X_X

Reply 25

Zacken

Original post by Quido

Hmm, I just tried it again using simultaneous equations.
I used s = ut + 1/2 at^2
I got 30 = 2u - 2a and 60 = 6u - 18a and then solved them to get a = -2.5 and u = 17.5 m/s. And then using v = u + at I found t = 7.

30 \neq 2(17.5) - 2(-2.5)

Reply 26

drandy76

Original post by Quido

before setting up second stage, remember that the final velocity at stage 1 is the same as the initial velocity at stage 2

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Reply 27

B_9710

I got V as 12.5ms^-1 and U as 17.5ms^-1 (V is the speed after 2 seocnds has passed from t=0). But yes T is 7s to come to rest.

(edited 8 years ago)

Reply 28

Quido

Original post by B_9710

I got U as 12.5ms^-1 and V as 17.5ms^-1 (V is the speed after 2 seocnds has passed from t=0). But yes T is 7s to come to rest.
I think you have used the right method but mixed up your U and V. Post workings? (sceenshot)

The train is decelerating so how can it be possible for the final velocity to be higher than the initial?

Reply 29

Quido

Original post by Zacken

30 \neq 2(17.5) - 2(-2.5)

Sorry, I meant 30 = 2u + 2a

Reply 30

B_9710

Original post by Quido

The train is decelerating so how can it be possible for the final velocity to be higher than the initial?

Other way round.

Reply 31

Quido

Original post by B_9710

Other way round.

But we know at the end V = 0 as the train is brought to rest.

Also can I have help on this question.

A ball is thrown vertically upwards at 25 m/s. Find the length of time for the ball is above 3m from the point of projection.

s = 3
u = 25
v = ?
a = 9.8
t = ?

I got v = 26.15 m/s using v^2 = u^2 + 2as and then got t = 0.117s which is time the ball is under 3m.
How would I go about working out the time the ball is above 3m from there?

Reply 32

B_9710

Original post by Quido

Use s=ut+1/2at² and you should form and solve the quadratic and get 2 values for t.Find the difference between the two t values and that is the required time T. Acceleration is acting in opposite direction to initial motion of the ball.

(For the previous question I did not make V - the final speed/velocity - where the train had stopped. Please read my post again).