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# M1 - Constant acceleration watch

1. (Original post by B_9710)
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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2. (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
sorry I am teaching all day
3. (Original post by TeeEm)
sorry I am teaching all day
thought so >.>
4. (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
Yep, you've done this one wrong - sorry. 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.
5. (Original post by Zacken)
Yep, you've done this one wrong - sorry. 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.
all these c2 papers making me compile all my knowledge is kill me X_X
6. (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.
7. (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.
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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8. 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.
9. (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?
10. (Original post by Zacken)
Sorry, I meant 30 = 2u + 2a
11. (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.
12. (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?
13. (Original post by Quido)
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?
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).

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