SUVAT Equations of Motion Questions - A Level Physics
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- เผยแพร่เมื่อ 10 เม.ย. 2015
- This video explains solving questions that involve the equations of motion (SUVAT equations) for A Level Physics.
Struggling with how to solve real problems with the suvat equations? This video shows you a technique that will guarantee you the correct answer every time.
Thanks for watching,
Lewis
This video is recommended for anyone studying A Level Physics in the following exam boards:
AQA
CIE
Edexcel
Edexcel IAL
Eduqas
IB
OCR A
OCR B
WJEC
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fast paced but easy to follow videos are hard to find on youtube. you show everything needed to know in a short amount of time and it's easily rewatchable
I love how you have divided topics into small videos covering the subtopics....... youve made phyisics for me so much easier
Best physics videos on youtube
Thank you, I try my best!
Awesome, been struggling with this in physics for a bit, very clear and helpful :) Thanks
Thankyou so much this helped a lot.
Had to watch this twice but really helped thanks
Really helped me. Thank you.
Really enjoy your videos, will you ever made some on electricity?
hahaha, I had to watch the video twice before I understood what the "Suvat" equation was.
They call it SUVAT because people commonly use those letters for the 5 variables of uniformly accelerated motion. S for distance, because d already has another full time job in Calculus, and u for initial velocity as the alphabet neighbor of v, in order to avoid subscripts.
Can you show how you'll work through more complex and involved calculations eg the 5 and 6 mark questions
Ayoola Oluwanusin I will do, working on it during the next week.
Thank you, much appreciated
Hi, I have a question. A girl is standing on a bridge and throws a coin with a vertical velocity of 5 ms^-1. It hits the water below the bridge after 1.5 seconds. Assuming that the effects of air resistance are negligible, what was the initial height of the coin above the water?
Did you figure it out? Its been 4 years? 😂
Sir please tell me I had a doubt in one question that is in physics can you tell me.
An object started from rest u=0 and covered 100 m of distance in 5sec if we measure it's speed that v=s/t then it will be 20 m/s but if we go through by equationsbof motion and by using second equation we can find acceleration and then by 1st we can find v but object is traveling in same direction but the velocity is coming 40 m/s it is half of the speed but why.In my all numericals this thing is happening
v=s/t just works out the average speed. If it is uniformly accelerating then it starts at 0m/s and ends at 40m/s as it continues to get faster.
can we use s=ut+1/2at2 instesd of finding v?
Yes that's what i did
Hi I wanted a to know which pens you use to write these I like the red and black that you use
Noel Clark They are Pilot V Sign pens
in the example would use also put acceleration as a negative value.
no, as in that particular situation downwards is the positive direction (as its the only direction the object does)
how do we know when to use acceleration values as 9.81 meter per seconds square
its gravity..
gravity on earth is always 9.81. its different on other planets tho
If it takes place on this planet, use 9.81 m/s^2 unless otherwise specified.
Hi can you help me with this question. It's a 1 marker and I feel I have done so much working out to solve it but I am not getting the answer. I clearly I am thinking too much when there is something so basic to it, it's part of the new spec questions.. A explanation on how to do it would be helpful.A car accelerates uniformly from rest along a level road. The effects of air resistance on the car are negligible. The car travels 12 m in the 2nd second of its journey.
How far does it travel in the fourth second?Thanks.
+Mahmzo Is it 24m/s?
In the 2nd second it travels with a velocity of 12m/s. If it was accelerating uniformly then it would have a velocity of 6m/s after 1 second, 12m/s after 2, 18m/s after three and 24m/s after 4 seconds.
The answer is actually 28 i was so confused
Mahmzo I'm just seeing this video now but this is how I did it if you still need help. You use suvat algebraically
Within the first second we know that u=0 t=1 a=a v=v
We can use v=u+at to find v in terms of a
V=0+a(1)
V=a
Within the second second we know that s=12 a=a u=a (from equation before) t=1
S=ut +1/2at^2
12 = a(1) + 1/2a(1)^2
12=a+1/2a
24=2a+a
24=3a
a=8
Using the acceleration we know in fourth second that u=24 v= 32 t=1 s=?
S=(u+v)/2 *t
S= (24+32)*1/2
S=28
Hope this helps!! :D
is this for the new spec too?
Yes it is
Thank you :)
How do you know if a question requires you to use SUVAT equations.
Jeff Asiamah when Acceleration is constant you use a SUVAT equation
one question, how do we know when gravity is 9.81 and when is it -9.81, I always get this wrong
Anytime it takes place on this planet, it is 9.81 N/kg.
As for the sign, it has to do with which direction you assign as positive. If you assume down is positive, then g is positive. If you assign up as positive, then it is negative. It's common for formulas to only refer to its magnitude, such as GPE=m*g*h, and W=m*g. You simply need to account for direction with context.
this might just be really stupid, but I don't understand why mass doesn't come in to these equations? Like for that problem there a mug would fall in the same amount of time as a really heavy weight?
These are kinematic quantities, where we don't care what causes the motion, we just care about mathematically describing it.
In the case of free fall specifically, because weight and inertia are a package deal that are both proportional to the mass, mass cancels out of the equation, and the acceleration is the same regardless of any of the properties of the object that is falling. That is, as long as air drag is negligible, which it is unless otherwise specified.
when there is no air resistance, mass doesnt matter. its hard to get ur head around it but if u dropped a brick and a feather at the same time in space (a vaccum) they'd both land at the same time
At the end you said to put it in 3sf but you put it in 4sf because you count the 0, so 0.639 is 4sf so the answer should be 0.63
actually it should be 0.64 haha i pressed the wrong number lol
+madeline McCann You don't count the leading zero as a significant figure.
As the smart guy said, Anything after a 0 counts as a significant figure such as: 0.00639 is still only 3 sf, but 0.609, that is also 3sf
im now more of a science