interpreting motion (1 Viewer)

[Masaken]

ok so this isn't probably even maths ext 2 but it's in the 4u textbook so i'll just put it here

how do i interpret this q? like i did part (a) and (c), but (b) just got me stumped. when the ball is rising, i know it's going up at a constant velocity but gravity (10 m/s^2) is also acting on it. how do i relate this back to the ideas of motion to explain why v = that expression??

 

[synthesisFR]

because u let upwards be positive
velocity has to be the positive component because ur going upwards

 

[Masaken]

because u let upwards be positive
velocity has to be the positive component because ur going upwards

wait what is that literally it

bruh the cambridge worked solutions went on a tirade about newton's first and second laws of motion

 

[s97127]

ok so this isn't probably even maths ext 2 but it's in the 4u textbook so i'll just put it here

how do i interpret this q? like i did part (a) and (c), but (b) just got me stumped. when the ball is rising, i know it's going up at a constant velocity but gravity (10 m/s^2) is also acting on it. how do i relate this back to the ideas of motion to explain why v = that expression??

View attachment 38554

Can you show me how you use x'' = -10? I know how to solve as a physics problem but not as 4u math problem. For question b, the ball will go up slowly until v reaches 0 at its max height of 20m and then it will go down after that. So when x <= 20, the ball still rises and 400 -20x >= 0 so v = sqr(400-20x)

 

[Masaken]

Can you show me how you use x'' = -10? I know how to solve as a physics problem but not as 4u math problem. For question b, the ball will go up slowly until v reaches 0 at its max height of 20m and then it will go down after that. So when x <= 20, the ball still rises and 400 -20x >= 0 so v = sqr(400-20x)

you can look at q11 as a projectile motion q i think, and since it's thrown vertically upwards that means the only force acting on it is gravity (hence x'' = -10 as gravity is the force acting on the ball in the opposite direction). but the way the exercise was structured it was asking me to apply the suvat equations which work here cos acceleration is constant (i believe suvat is mostly used in physics though) - i used the v^2 = u^2 +2as one, with u = 20 and a = 10 (and let s is equal to displacement x) --> so you sub that in and you end up getting the result in (a)

(someone pls correct me if i'm wrong, not really good at explaining motion)

 

[unofficiallyred12]

q11a it's intended for u to use the fact that a = dv/dt = dv/dx x dx/dt = vdv/dx (or alternate form which is d/dx(1/2v^2) since suvat equations are non-assumable in 4u

 

[Average Boreduser]

q11a it's intended for u to use the fact that a = dv/dt = dv/dx x dx/dt = vdv/dx (or alternate form which is d/dx(1/2v^2) since suvat equations are non-assumable in 4u

Yooooo asian 5 subject selection

 

[Masaken]

q11a it's intended for u to use the fact that a = dv/dt = dv/dx x dx/dt = vdv/dx (or alternate form which is d/dx(1/2v^2) since suvat equations are non-assumable in 4u

yeah usually that's what you do but then when i went to mark it they used suvat (the previous q was about proving a suvat eq) so i did the suvat way as well, but yea @s97127 don't use suvat in 4u lol the textbook is just quirky

 

[s97127]

you can look at q11 as a projectile motion q i think, and since it's thrown vertically upwards that means the only force acting on it is gravity (hence x'' = -10 as gravity is the force acting on the ball in the opposite direction). but the way the exercise was structured it was asking me to apply the suvat equations which work here cos acceleration is constant (i believe suvat is mostly used in physics though) - i used the v^2 = u^2 +2as one, with u = 20 and a = 10 (and let s is equal to displacement x) --> so you sub that in and you end up getting the result in (a)

(someone pls correct me if i'm wrong, not really good at explaining motion)

I don't have the textbook but i think i get it now. Thanks