04 - Motion with Constant Acceleration Physics Problems, Part 2

Learn how to solve physics problems that involve equations of motion with constant acceleration. This problem involves driving with a constant acceleration and avoiding an obstacle.

We must first decide which equation of motion is useful to solve this problem, then draw a diagram and solve the problems using algebra.

[lythia]

IM ONLY 15 & COMPLETELY INLOVE W/ PHYSICS NOW!!!THANKYOU SO MUCH I CANT EVEN EXPLAIN HOW EXCITED I AM!!!!ITS LIKE WHEN U FIND A BEAM OF LIGHT THAT LEASE U OUT OF A DARK CAVE! MY MIND IS COMPLETELY BLOWN OFFFF

[Frank-sijd]

Thanks for your very clear math explanations! Your way of teaching should be standard for math

[juanibanez]

Thank you. Best teacher ever. I stopped school because of pressure and became a proffesional martial artist now I crave all sciences and you are a blessing so easy to understand I rewatch your videos over and over and try to master them.

[Lasseu]

Thank you for this, got test in few weeks about this subject. This is the same one that got past me and now i get it :)

[LONE-MUHAIB]

Thanks sir, most impressive professor in my opinion.

[ropamoses]

I like the way you teach. You're the best physics teacher .

[michaelemmanuel]

I was almost confused on this problem. Thank God you used another method in solving it

[strugglingstudent]

thank you for all your support, wish all teachers taught this clear

[yorps]

what about if u balance the object without hitting it

[AVLife]

Wait, wouldn’t he have hit the obstacle because we used 10m as the Vo instead of 0m? If 43.3 meters was used up when the deceleration started wouldn’t that be an Xf of 53.3 meters?

[yaseerkadam]

I need more explanation about the second part (Time needed to slow down), where did we get -20? And i divided the numbers by -6 but i didn't get 3.33.

[israelo]

You are great teacher! I wish I had you as my teacher. You look so easy going...

[francafiandri]

Thank you a lot! You're so clear!!

[testplmnb]

Dafook?
Speed 20m/s
Distance: 50m
How long it takes to cover the distance: 2.5 sec

How long it does to deaccelerate from 20 to 0: 3.33 secs which is extended by 0.5 secs of delay.

3.83 secs vs 2.5 second?

[realitybites]

I like the longer formula because it is more concise.

[SyedTayyabAli-rywb]

Hello, before car starts to deccelerate they have covered 10m. so remaining distance is 40m. after applying the break it starts deccelerating and stops after 43.33m so, i am confused, wouldnt they collide ? pls explain

[theshoktron]

For the first equation you could have used velocity times time and that gives you 10

[kallisingh]

A car traveling at 20m/s will take 2.5 s to cover 50m. The time it takes to stop is 3.33s (according to your calculation) whether from initial point 0m or 10m after reaction time clearly tells me he will collide. I worked it mentally and conclude it will hit before you solved it. Consider also that the brake pedal will take a fraction of time to go right down so as to then get max deceleration. Please explain. The only part I got also was 10m for this distance he would cover after 0.5 sec reaction time when traveling at 20m/s.

[samaraellis]

This really helped me a lot thank you😊

[TheFarmanimalfriend]

I think the initial X should be marked as 0 m, not, 10 m. Acceleration does not start until then. It makes no difference in the outcome.