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# suvat

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1. a couple of question involving Suvat.....

"1. A car, moving with uniform acceleration along a straight level road, passed points A and B when moving with speed and respectively. Find the speed of the car at the instant it passed C, the mid-point of AB."

i thought C, being midpoint you could just find the average of the two velocitys given.

so, (30+60)/2 = 45ms^-1

however, the answer is 47.4ms^-1, i do have an assumption that it is a typo though, just need comfirmation.

"2. A car is moving along a straight horizontal road at a constant speed . at the instant when the car passed a lay-by, a motor-cyclist leaves the lay-by, starting from rest, and moves with constant acceleration in pursuit of the car. Given that the motor-cyclist overtakes the car T seconds after leaving the lay-by, calculate:

a, the value of T
b, the speed of the motor-cyclist at the instant of passing the car."

i really dont understand how this can be drawn out..mainly because there are 2 vehicles and start-stop, is this suppose to be modelled using a speed-time graph?
2. For part 1. the book is correct.

If the velocity were increasing uniformily and you were asked to find the velocity when it had been travelling for half the time you would be correct.

However, it is asking the velocity at half the distance.

See spoiler for how to do first part.

Spoiler:
Show

Use V^2=u^2+2as for the whole journey A to B.
Assuming it start at A. This will give you "as".
Repeat the process for the half way point, substituting in your value of "as".

3. (Original post by KeineHeldenMehr)
i really dont understand how this can be drawn out..
2. a) For motor-cyclist, u=0 a=2.5 t=T s=?. For car, s=18xT. Use simultaneous equations.
2. b) You will have t so find v for motor-cyclist.
4. (Original post by ghostwalker)
For part 1. the book is correct.

If the velocity were increasing uniformily and you were asked to find the velocity when it had been travelling for half the time you would be correct.

However, it is asking the velocity at half the distance.

See spoiler for how to do first part.

Spoiler:
Show

Use V^2=u^2+2as for the whole journey A to B.
Assuming it start at A. This will give you "as".
Repeat the process for the half way point, substituting in your value of "as".

s=?
u30
v=60
a=?
t=?

hmm...
Repeat the process for the half way point, substituting in your value of "as"

yeah, i'm stumped.
5. For the second part the distance is half AB.
So, as/2

or more fully
6. (Original post by ghostwalker)
For the second part the distance is half AB.
So, as/2
ah gotcha~

=47.4....ms^-1.

thank you.
7. (Original post by OllyThePhilosopher)
2. a) For motor-cyclist, u=0 a=2.5 t=T s=?. For car, s=18xT. Use simultaneous equations.
2. b) You will have t so find v for motor-cyclist.
that worked wonders! thank you!
8. Generally you need three unknowns for this to work. In the first one it works because you are halving an initial distance. For the second part you know that t and s are the same for both. When they meet, if time 0 is when the car passes, they will have traveled the same distance and time.

Work out s in terms of t for motorbike. For the car just use v=s/t

See if you can get it from there.

Although generally, these seem harder than the typical exam questions I got.
9. ^thanks.

another question which i think i have solved however, little understanding to what is going on.
"A particle P moved in a straight line with constant retardation. At the instants when P passed through the points A, B and C it was moving with speeds: respectively.

prove that: .

going back to the first question in this thread, i use the same method "ghostwalker" has suggested....working out in terms of "as" for AB and then BC.

AB:
s=?, u=10,v=7, a=?, t=?

(-25.5)

BC:
s=?, u=7,v=3,a=?,t=?

so,

.

im just unsure why this can be done in terms of 'as'
surely as/as equals 1, so,

i know i dont make sense, but this is because i cant make sense out of this:

.

what happens to the 1?
10. You're using different values of s each time.
To clarify call the distance AB "b", and BC "c", then you will get values for "ab" and "ac".
When you divide you get b/c = (AB)/(BC)
11. (Original post by ghostwalker)
You're using different values of s each time.
To clarify call the distance AB "b", and BC "c", then you will get values for "ab" and "ac".
When you divide you get b/c = (AB)/(BC)
why did you clarify in that way though?
12. Not sure what you mean by that question.

The letter s represents two different things in your workings, so dividing "as" where s means the distanc AB by "as" where s means the distance BC is what I assumed you saw as confusing.

Hence my suggestion to use b, and c.
If you call the first distance b, and the second one c, you don't end up with (as)/(as).

They don't have to be b and c, just any two letters.
13. (Original post by ghostwalker)
Not sure what you mean by that question.

The letter s represents two different things in your workings, so dividing "as" where s means the distanc AB by "as" where s means the distance BC is what I assumed you saw as confusing.

Hence my suggestion to use b, and c.
If you call the first distance b, and the second one c, you don't end up with (as)/(as).

They don't have to be b and c, just any two letters.
ok, thanks....
14. Alternatively, when you get the result of the first part you can divide through by a, and say distance AB = -25.5/a.
Similarly distance BC = -20/a.
15. (Original post by ghostwalker)
Alternatively, when you get the result of the first part you can divide through by a, and say distance AB = -25.5/a.
Similarly distance BC = -20/a.
yeah that works,

s/s, a/a = 1

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