Bubble's specialist questions

[Mao]

an alternative method [that is slightly easier]:

the ratio of acceleration is 1:4, that implies both distance and time will be 4:1, or a fifth of the total [draw the v-t graph to convince yourself]

this means, the last 100m took 24s [a fifth of the total]





since "u" in this case (initial velocity) is the maximum velocity, and v (final velocity) is 0:



using ratios can save a lot of effort, and avoid large numbers such as those coblin encountered. however, it can get tricky as only in some cases can this method be used, and in most cases it cannot.
for example, where the v-t graph plateaus at a maximum velocity

[xox.happy1.xox]

This is the maths I like!

[bubbles]

This is the maths I like!

LOL

[bubbles]

I was heading in the right direction until i got stuck at step 2- finding the velocity equation.

The second one is:

All you did here was add the velocity from t= 0->96 and t= 96->120 to find the total velocity, is that correct?

an alternative method [that is slightly easier]:

the ratio of acceleration is 1:4, that implies both distance and time will be 4:1, or a fifth of the total [draw the v-t graph to convince yourself]

this means, the last 100m took 24s [a fifth of the total]

I need to brush up on my ratios -"...Mao could you explain why both distance and time will be "4:1" instead of 1:4.

[Collin Li]

All you did here was add the velocity from t= 0->96 and t= 96->120 to find the total velocity, is that correct?

Yes sort of. I added the "velocity gained," which is why I multiplied the time by acceleration (it's constant, so I can just go ).

[Mao]

an alternative method [that is slightly easier]:

the ratio of acceleration is 1:4, that implies both distance and time will be 4:1, or a fifth of the total [draw the v-t graph to convince yourself]

this means, the last 100m took 24s [a fifth of the total]

I need to brush up on my ratios -"...Mao could you explain why both distance and time will be "4:1" instead of 1:4.

the relationship between velocity, time and acceleration is

rearranging: , assuming velocity(s) is constant, we will end up with , an inverse relationship.
hence, a 1:4 ratio of acceleration will become 4:1

now, we also know that , a direct variation
hence, a ratio of 4:1 of t will imply a ratio of 4:1 for distance.

you can also think about this in terms of what happens in real life: if you do someting 4 times as fast, you would take only a quarter of the time. 1:4 --> 1:0.25 == 4:1

[bubbles]

Thanks I get it now 

[bubbles]

Other expressions for acceleration
A particle moves in a horizontal line so that the velocity, v m/s, is given by . Find:
a) the position, x m, in terms of time t seconds, given that, when t=0, x=1








t=o, x=1




... which is wrong --"
the correct answer is:


(hmm on second thoughts.. are both answers correct. Is it one of those ? If this is the case, which is the preferred form in the exam, does it matter?

[shinny]

Yeh, both are correct. It's a pretty basic trig identity and you're expected to know it. In the case of an exam, I'd say simplify it to x=cos2t incase u get a pedantic examiner but all in all, i HEAVILY doubt you'll ever lose a mark for not simplifying it.

[bubbles]

Resolution of forces and inclined planes
A particle of mass 5kg is being pulled up a slope inclined at 30° to the horizontal. The pulling force, F newtons, acts parallel to the slope, as does the resistance with a magnitude of one-fifth of the magnitude of the normal reaction.
a) Find the value of F, such that the acceleration is up the slope.
F - (Fr + 5gsin30) Fnet
R = 5gcos30
Fr = (1/5)(5gcos30)
Fr = gcos30
F = 5(1.5) + gcos30 + 5gsin30
F= 40.49 N

b) Also find the magnitude of the acceleration if this pulling force now acts at an angle of 30° to the slope (i.e. at 60° to the horizontal).

How do you do b) ?

[shinny]