suvat

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

Reply 2

OllyThePhilosopher

11

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.

Reply 3

OP

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

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

v^{2}=u^{2}+2as

60^{2}=3=^{2}+2as

2as=2700

as=1350

hmm...

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

v^{2}=u^{2}+2(1350)

yeah, i'm stumped.

Reply 4

For the second part the distance is half AB.
So, as/2

or more fully

v^2=u^2+2\frac{as}{2}

Reply 5

OP

ghostwalker

For the second part the distance is half AB.
So, as/2

ah gotcha~

v= \sqrt{30^{2}+1350}

=47.4....ms^-1.

thank you.

Reply 6

OP

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!

Reply 7

abstraction98

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. :smile:

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

Reply 8

OP

^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:

10ms^{-1}, 7ms^{-1}, 3ms^{-1}

respectively.

prove that:

\frac{AB}{BC}=\frac{51}{40}

.

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=?

v^{2}=u^{2}+2as

as=\frac{-51}{2}

(-25.5)

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

v^{2}=u^{2}+2as

as=\frac{-40}{2}

so,

\frac{-51}{2} \div \frac{-40}{2}

=\frac{AB}{BC}=\frac{51}{40}

.

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:

\frac{AB}{BC}=\frac{as}{as}=\frac{\frac{-51}{2}}{\frac{-40}{2}}

1=\frac{51}{40}

.

what happens to the 1?

Reply 9

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)

Reply 10

OP

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?
:confused:

Reply 11

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.

Reply 12

OP