STAV 2010 problem

[kenhung123]

For this question I found the area under the graph to be 30x1000x2=60000
Then I equated that value to 1/2mv^2 to find v. Is that wrong?

[kenhung123]

Also this question I found time by t=x/v=22/4.5=4.89
u=0 (solving with vertical component
v=?
a=-10
t=4.39


Turned out v=48.89m/s

Then impulse=mv=88x48.89=4302=Ft
F=4302/48.89=880N

[TrueTears]

For this question I found the area under the graph to be 30x1000x2=60000
Then I equated that value to 1/2mv^2 to find v. Is that wrong?

I'd do it this way

f * (change in time) = I = m * (change in v)

60000=2200(change in v)

[kenhung123]

Force distance graph can be used to find momentum?

[cipherpol]

For this question I found the area under the graph to be 30x1000x2=60000
Then I equated that value to 1/2mv^2 to find v. Is that wrong?

Units for the area under graph is Ns which is impulse, not kinetic energy.

[m@tty]

Force distance graph can be used to find momentum?

It isn't force-distance, it is force-time. The area under a force-distance graph(F x d; work) is energy, but the area under a force-time graph(N x s) is impulse or change in momentum.

[kenhung123]

AHhh silly me! Thanks for clarification. So how about the projectile question?

[m@tty]

Vertically he traveled only 3 meters in 4.89s which means there must have been some upwards force countering gravity - Lift.

a=?  s=3 m  u=0 m/s  t=4.89 s  v=?





down.

[kenhung123]

Umm I thought in VCE we don't deal with air resistance in projectile :S

[m@tty]

In this question they have specifically stated that this man was trying to fly... For this question you have to take it into account vertically but at the same time ignore it horizontally.

[ghadz7]

27 m/s ??

[Greggler]

fat mav

[Richiie]

Doesn't this question ask for the velocity BEFORE braking? This question has stumped me. Is the answer what ghadsz7 said?

[m@tty]

27 m/s ??

Yep, 27.3 m/s.

[kenhung123]

Answers like 13.6