Physics: 2 Kinematics question help
-
Q1. "A car is travelling with uniform acceleration along a straight road. The road has marker posts every 100m. When the car passes one post, it has a speed on 10m/s and when it passes the next post, its speed is 20m/s. What is the car's acceleration?"
Q2. "A passenger is sitting in an aeroplane, which takes off and climbs to 10km. During this time, what happens to the weight of the passenger?"
A1. 1.5m/s^2.
A2. Weight decreases. [Does it mean that gravitation force decreases since W=mg?]
Help greatly appreciated, thx!
-
dear hewhoshallnotbenamed,
for question 2 .
i looking at the question is not properly set.they shouldnt have ask what happens to the weight of the passenger ,because its ambiguous as i will assume as when the plane completed climbing to 10km .i would say the answer is correct .but NOT cambridge way of setting question [prone to get criticism from local examination syndicate]
University of Cambridge would never set a question so prone to attack on the language and ambiguity
-
Q1.
v^2 - u^2 = 2as ...(1)
a = (v^2 - u^2)/(2s) = (20^2-10^2)/(2*100)=1.5m/s^2
In case you did not learn (1) in school, you should have learnt the following:
(v+u)/2 * t = s ...(2)
(v-u)/t = a ...(3)
The product of (2) and (3) give your the result in (1).
Q2.
The weight decreases in the process of climbing up. (it is a continuous process, as g decreases continuously).
-
oh i see! i only learned the 3rd equation. thanks a lot!
-
hewhoshallnotbenamed, are u sure that your answer to the second question is right ? I do not think that weight will decrease, the apparent weight will increase. The increase in height(i.e 10km) is insignificant compared to the radius of the earth. therefore the change in gravitational acceleration(g) is negligible. Looking at the qn, i think you are probably a secondary school student, so i shall not explain my previous statement in depth by talking abt Newton's law of gravitation. Since you are a sec school student, you have also learnt that g is constant in earth. Before i answer the qn, it is important for you to understand the term apparent weight. As the name suggest apparent weight is not your actual weight. Your actual weight actually doesn't change. When you stand on a weighing scale, the reading that you see on the weighing scale is actually the normal force(N). Since you are stationary, N=mg, so you can read off your weight. Ok now lets say you are the one who is in the aeroplane. When the aeroplane takes off and goes up 10 km, the aeroplane is accelerating. Since you are in the plane, you also experience the acceleration(a). So the resultant force,as your acceleration is upwards, will also be upwards. So what are forces is acting on your body. Normal reaction force(upward) that is acting on your body and your weight(downward). therefore, ma=N-mg. Since i have explain earlier that N=apparent weight and in this case as you can see N=ma+mg=m(a+g), which means your "weight" has actually increase. Please ignore my answer if my answer is wrong and the answer is actually as simple as it is explained on the preceding post. but you have learnt something new.
-
hi plax, thx for ur (really long and in depth) response.
the answer given was that from the answer key. what my answer was was that the weight increases. how i got my answer was similar to the weighing scale explanation u had given. it was an MCQ and it only had the options of weight increasing or decreasing.
yes im a sec 4 student and my physics o lvl MCQ is on friday. oh dear.
-
I strongly disagree.
1. Climbing up to 10km does not neccesarily imply 'accelerating'. In fact, only an intial acceleration is needed, after that, the speed can remain at constant value, or even decreases just before it reaches 10km, so the assumption that there is an acceleration throughout is invalid.
2. Moreover, there is no problem with the fact that, the weighing scale shows a larger reading when the plane accelerates upwards, as the weighing scale is measuring the pressing force (or the magnitude of the normal reaction force). However, your weight is not affected by this factor, it just means, the weighing scale on an accelerating plane, will not show a reading which is equal to your weight. Weight refers to the gravitational pull exerted on the object due to the presence of a gravatational field, in this case, generated by the Earth, so it is independent of the motion of the plane.
3. 10km insignificant comparing to the radius of the Earth, 6000+km (0.17% approximately). However, it does not mean the change in g value is also insignificant. 0.17% of 9.81 is approxmately 0.02(2d.p), which implies the g will be decreasing from 9.81 to 9.79 continuously in the process. Therefore, the weight decreases. In fact, it is not logical to say that a change from 9.81 to 9.79 shall not be regarded as a decrease but a change from 9.81 to 9 or smaller shall be.
-
Originally posted by frekiwang:
I strongly disagree.
1. Climbing up to 10km does not neccesarily imply 'accelerating'. In fact, only an intial acceleration is needed, after that, the speed can remain at constant value, or even decreases just before it reaches 10km, so the assumption that there is an acceleration throughout is invalid.
2. Moreover, there is no problem with the fact that, the weighing scale shows a larger reading when the plane accelerates upwards, as the weighing scale is measuring the pressing force (or the magnitude of the normal reaction force). However, your weight is not affected by this factor, it just means, the weighing scale on an accelerating plane, will not show a reading which is equal to your weight. Weight refers to the gravitational pull exerted on the object due to the presence of a gravatational field, in this case, generated by the Earth, so it is independent of the motion of the plane.
3. 10km insignificant comparing to the radius of the Earth, 6000+km (0.17% approximately). However, it does not mean the change in g value is also insignificant. 0.17% of 9.81 is approxmately 0.02(2d.p), which implies the g will be decreasing from 9.81 to 9.79 continuously in the process. Therefore, the weight decreases. In fact, it is not logical to say that a change from 9.81 to 9.79 shall not be regarded as a decrease but a change from 9.81 to 9 or smaller shall be.
Wow, this is an O level question, don't need to be so serious...
Seriously, this question is just cock, so ambiguous
-
its one of a 1000 mcq questions in the revision package my school gave me. i think this question is from another school's prelim.
i agree this question is quite unclear.
anyway, i appreciate the help u guys have offered. thanks!
-
frekiwang, chill!!!!
-
I never felt comfortable with using the kinematical formulas to solve kinematics questions. I provide an alternative graphical method for question 1:
While the use of the graphical method in this question seems trivial, it can turn out to be more efficient in certain questions.
Refer, for example, to my website.
-
At O levels, most students do not learn kinematics equations apart from a = (v-u)/t
Some schools do teach, but it is mostly unnecessary.
The graphical method is the 'way' that is taught at O levels :)
-
I am trying to 'trvialize' Physics to just a few fundamental principles.
Everything can be derived from these few principles. That's what makes Physics appealing to the physicists!
In the case of Kinematics, everything can be boiled down to just two equations:
1. a=dv/dt
2. v=ds/dt
I never understood why schools are teaching v^2=u^2+2as, etc.
-
Hi,
v^2=u^2+2as is a simplified formula. In my time, my school taught the derivations of these formulas from graphs, which makes me appreciate its usage.
Basically, v^2=u^2+2as is derived from substituting t = (v-u)/a into s = (u+v)t/2, of which both equations can be derived from a graph. The basic assumption is that acceleration is a constant.
That said, I also don't understand why schools are teaching and making subjects more difficult and complicated than it really is :)
-
Hi,
We simpletons should unite, haha :P
Cheers,
Wen Shih