Re: Few questions. (1 Viewer)

[BlackJack]

1. A vehicle is travelling at 9 ms.-1 due East and Observes a train travelling 12ms.-1 due South. What is the velocity of the train relative to the vehicle ?

Relative to the vehicle, you need to reverse the velocity of the observer, which is the vehicle. (i.e. relative to the vehicle, everything that stands still is travelling 9 ms.-1 due West already) Then add tail to head in whatever order. You derive veocity = sqrt( 9^2 + 12^2 ) = 15 ms.-1 south-west.

3. A 1600kg Vehicle travelling at 14.5ms-1 is slowed to a half voer a distance of 350m. Determine the combined total of the frictional forces that slow the vehicle down.
Please include all working out - Greatly appreciated!

the Work-Energy relation thingy, assuming that friction forces are constant throughout travel (do you still have this simplification in highschool?):

Work done = force * distance = change in kinetic energy.
.'. W = F*350 = 0.5m v_f² - 0.5m v_i².
= 0.5m (v_f² - v_i²).

So,
F = - 0.5 * m / 350 * ( v_i² - 0.25*v_i²
= - 0.5 * 1600 / 350 * 0.75 * v_i².
~ 360.4 N against the direction of travel.

 

[tommykins]

Few questions.

These are from my half yearly and I'm using them as revision - some of the questions I simply don't get how the answer is derived.

1. A vehicle is travelling at 9 ms.-1 due East and Observes a train travelling 12ms.-1 due South. What is the velocity of the train relative to the vehicle ?

---> 9m/s
|
|
v 12m/s


I know you're supposed to add tail to head, but I don't know which way? I found the velocity value, just not the direction, can someone point me in the right direction? Could you also point out how to do questions related to this? I'm a tad confused.

2. A 1500 kg vehicle travelling 3ms.-1 west colldies witha 1200kg vehicle travelling 7m.s.-1 east. The collision takes 0.12 s to occur. Determine the average force of the 1200kg vehicle on the 1500kg vehicle.

3. A 1600kg Vehicle travelling at 14.5ms-1 is slowed to a half voer a distance of 350m. Determine the combined total of the frictional forces that slow the vehicle down.

Please include all working out - Greatly appreciated!

 

[z600]

1) reverse the vector of the train and add it to the vehicle. You dont change the observer

2)P (vehicle 1)=1500 x 3
=4500

p(vehicle 2) = 1200 x (-7)
=-8400

change of momentum is 4500--8400 = 12900

impulse =Fx delta t

12900=F x 0.12

F=107500N

LOL am i right here?

3)3. A 1600kg Vehicle travelling at 14.5ms-1 is slowed to a half voer a distance of 350m. Determine the combined total of the frictional forces that slow the vehicle down

i will come back and do it XD

 

[helen888]

In relation to the second question, I don't understand why Blackjack had:

" F = - 0.5 * m / 350 * ( v_i² - 0.25*v_i²
= - 0.5 * 1600 / 350 * 0.75 * v_i²."

if velocity slowed to half the initial velocity, wouldn't it be:


F = 0.5*m/s * (v²-u²)
=0.5*1600/350 * (7.25²- 14.5²)
= 0.5*1600/350*-157.6875
=-360N ?


Same figure in the end, but I'm confused about your working out...

 

[alcalder]

These are from my half yearly and I'm using them as revision - some of the questions I simply don't get how the answer is derived.

1. A vehicle is travelling at 9 ms.-1 due East and Observes a train travelling 12ms.-1 due South. What is the velocity of the train relative to the vehicle ?

---> 9m/s
|
|
v 12m/s


I know you're supposed to add tail to head, but I don't know which way? I found the velocity value, just not the direction, can someone point me in the right direction? Could you also point out how to do questions related to this? I'm a tad confused.

Another thing you can do to work it out is imagine you are in the vehicle. OK, now you are travelling EAST, the train is travelling SOUTH away from you. Now, if you can, imagine that the vehicle is not moving at all. What way is the train going. Well it is going in a south westerly direction. Now, change your vectors to reflect that.

It says to me that you should be reversing the vector of the vehicle (not the train).

If you reverse the train vector then the relative motion of train appears that it is going in a north easterly direction - uh uh, no it isn't. But if you were sitting on the train, sure the vehicle would look as though it is going in the north easterly direction.

You are allowed to use some common sense to get to your answer. So, shut your eyes and imagine yourself in the situation, then do the maths. I was always taught one more step when doing Physics questions. Ask yourself "Does the answer make sense?" and to do this you have to really think outside the square a little rather than just about the numbers.

Hope that helps. Vectors suck if you have never done them before.