Acceleration Time Graphs Area Kinematics Physics Tutorial

[ReviRiabinski]

Crazy how this video was published five years ago... Anyways, I'd like to thank you for the clear instructions as i have to watch your videos everynight since my teacher fails to teach properly. Thanks 

[TokyoJuice]

Thank you so much for these videos! I just started 11 physics and I think just knowing it's going to be tough scares me a little. It's hard to understand what my teacher is trying to convey sometimes, and these videos fill up all the gaps!

[PhysicsEH]

@JDD6214 Hi. We don't graph that ever but I would think you should draw it as a series of flat lines (like a staircase) because you have data for different intervals of time such as the acceleration from 2 to 4 seconds, so you should draw a horizontal line from for the acceleration for 2 to 4s. I would check with your teacher on this one.

[swift-za]

nah imma comment again guys....this guy is a GIFT from God❤️

[mosawo]

learning from a video posted 12y ago was really helpful

[PhysicsEH]

@AKN0111 Kart A uses v=d/t (uniform motion) so t=d/v=1000/20=50s Kart B uses d=V1t+1/2at^2 so 1000= 0 =1/2(0.333)t^2 solve t=77s So kart A wins in 50s. Where is kart B at 50s? use d=V1t+1/2at^2 and get d=0+1/2(0.333)(50^2) = 832.5m So 1000 - 832.5 = 167.5m behind. I hope you were able to follow this.

[PhysicsEH]

@videopost987 Sounds like you got it right. I didn't go into too much detail but if the velocity was 2 m/s at the start then the area tells us that the velocity increased by 6 m/s to a total of 8m/s at 2 seconds then it decreased by 4 m/s by the time is got to the 4 seconds so the velocity was 8-4= 4m/s at that time. I hope this helps.

[HannaLimisaCOhort]

Hi Sir, this video was vey helpful for me as someome who but is quite weak at it. I have one question which I hope you would answer. According to your graph, is the area under the accelration-time graph equal to the velocity? Thank you

[HannaLimisaCOhort]

Hello sir. This video has helped me alot to understand acceleration-time graphs. However I am not very good at physics, and cant grasp concepts quickly like my school mates. I have one question which I hope you could reply to: is the area under the acceleration-time graph equal to the velocity? Thank you sir and I'm sorry for asking such a stupid question.

[mmartinjosefsson]

You forgot the negative sign when you wrote the final answer for the negative area part ;)

Thank you for this tutorial though, helped a lot!

[WinterOnASolstice]

Thanks so much for the video! You helped clarify this A LOT.

[jannatulnyeemratul]

Sir, I would be very grateful if you answer this question: How is that a object accelerating at a speed of 9.8m/s but just covering 4.9m?Hope to get a big explanation.

I am having a lot of trouble with this questions.So please kindly reply, Sir.

[PhysicsEH]

@jaffarm They work...unlike my computer. Glad to hlep.

[hjhardscopez]

Whats happening at t=2 that straight line going down?

[AKN]

hi. i'm in grade 11 and we've just started kinematics and im already confused but ur videos are extremely helpful, but a had a question that im not understanding, hope u could help.
two go-carts, A and B, race each other around a 1 km track. Go kart A travels at a constant speed of 20 m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333 m/s^2. Which go-cart wins the race and by how much?

[HappyDaysRUs]

Thank you so much! Very helpful. Liked and subscribed!

[gonengg]

can't you find the total distance travelled (assuming the object was as rest before acceleration)?

[albdim]

I generally love your videos but in this case I think the use of West and East is wrong. The explanation should be that during the first 2 sec the speed was increased with 6 m/sec (from say 10 to 16 m/sec) whil eduring the next 2 sec the speed decreased by 4 m7sec (16 -4=12 m/sec). So, it is not the dirction of speed which changes, but while in the first part the speed increases, in the second part it decrease.

[videopost]

So is the graph stating that the unknown speed increase in velocity of 6 m/s then decrease by 4m/s. Anyway thanks a lot.