There is no such rule that an acceleration makes a speed time graph a straight line. In fact, it is almost impossible irl for a uniform acceleration, due to many factors such as gravity, air drag, friction etc. You might be a little misleaded on this, as this graph only states that the velocity is changing in a not uniform rate, but as the graph starts from 0 and moves upwards, it can be assumed that it is in fact accelerating
no 4 motion OP is a curve,and its gradient is positive(so,accelerates). Just looking at the curve, as the time increases ,the velocity increases,how can it be a deceleration?
According to this post Can I have some help with this? http://iob.imgur.com/uYGg/M0YkKp8cxy many agree that the answer for number 4 is C. Could you provide an explanation as to why you think the answer is D? Still, thanks for the post! Much appreciated
nope，from question，we need to produce a big,upright(virtual) image,if the answer is B,the image will be a upright(virtual) but deminished image,still D is the correct answer
hv sc kertas1 answer？
ReplyDeleteCould you please recheck the answer for question 4. Some said the answer is C
ReplyDeleteAiril izzah the answer for the fourth question is D.. Refer the book
ReplyDeleteMay I know why it's D??? I've put that as my answer tho but quite a number of ppl also say C ..:(
ReplyDeletefor question 4,motion at OP is not decreasing acceleration?
ReplyDeleteno 4 OP is a curve graph, if acceleration must be exact straight line, curve is deceleration
ReplyDeleteAccording to success book, it's still acceleration, just not uniform.
Deleteno matter it's a straight line or curve,if the gradient is positive,it's acceleration,their difference is just uniform and ununiform acceleration
DeleteThere is no such rule that an acceleration makes a speed time graph a straight line. In fact, it is almost impossible irl for a uniform acceleration, due to many factors such as gravity, air drag, friction etc. You might be a little misleaded on this, as this graph only states that the velocity is changing in a not uniform rate, but as the graph starts from 0 and moves upwards, it can be assumed that it is in fact accelerating
Deleteim sorry but that is accelerataion-time graph right?The question is velocity-time graph
DeleteBut still quite a lot of people say is C. Their reason is according to success, it says accelerate to achieve maximum velocity
ReplyDeleteHave science paper 1 ans?
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteno 4 motion OP is a curve,and its gradient is positive(so,accelerates).
ReplyDeleteJust looking at the curve, as the time increases ,the velocity increases,how can it be a deceleration?
According to this post Can I have some help with this? http://iob.imgur.com/uYGg/M0YkKp8cxy many agree that the answer for number 4 is C. Could you provide an explanation as to why you think the answer is D? Still, thanks for the post! Much appreciated
ReplyDeleteCn i get science...?
ReplyDeleteCn i get science...?
ReplyDeleteTo be honest, regarding on Question 4.
ReplyDeleteIt is testing your knowledge on terminal velocity.
As the gradient is positive, it is called acceleration.
As the gradient is negative, it is called deceleration.
Regarding my comments in this question, I would say a badly made question and should be removed and not considered in digital marking.
No. 23 is convex, isn't it? It's B
ReplyDeletenope，from question，we need to produce a big,upright(virtual) image,if the answer is B,the image will be a upright(virtual) but deminished image,still D is the correct answer
DeleteI agree with you
Deletecan someone explain to me how to get answer for no. 10? Thank you.
ReplyDeleteUse the formula infront which is s=ut+1/2at^2 oh and btw the english translation of the question is wrong in the book,refer the malay translation
ReplyDelete