Solving Acceleration Problems with Variables Explained

[Firebird]

Hello!

in Physics we just started acceleration

We were givin 3 equations

v=V(not) + at

x=V(not)t + 1/2at(Squared)

V(Squared) =V(not) + 2ax


Can someone please explain the variables and how to figure them out.

With those equations I need to figure problems like this

The speed limit in a school zone is 44 km/h. A driver traveling at this speed sees a child run onto the road 11 m ahead of his car. He applies the brakes, and the car decelerates at a uniform rate of 10.5 m/s2. If the driver's reaction time is 0.65 s, will the car stop before hitting the child?

YES OR NO
What is the stopping distance of the vehicle in m?


And


A boy throws a stone straight upward with an initial speed of 18 m/s. What maximum height will the stone reach before falling back down



I really don't know where to start. I normally start listing the varables like
V=
a=
t=
x=

But I do not understand what numbers to put in which areas,


Thank you very much

 

[CFDFEAGURU]

Well, to help you out a bit.

x is a distance in meters
t is a time in seconds
V is a velocity in meters/second
a is an acceleration in meters/second squared.

Also, V(not) for this problem is the initial velocity given.

Think it through, this is a very simple problem.

Thanks
Matt

 

[Firebird]

Thank you for your help Matt.

How do I figure out time, if I only get 3/5 or 2/4 varibles?

 

[CFDFEAGURU]

You are given the time. t = 0.65 seconds. I think you are misreading the problem. You are not calculating a time, you are calculating a distance. You need to determine if the driver stops within 11 meters. Not the time required for the driver to stop.

Make sense?

Thanks
Matt

 

[Firebird]

You are given the time. t = 0.65 seconds. I think you are misreading the problem. You are not calculating a time, you are calculating a distance. You need to determine if the driver stops within 11 meters. Not the time required for the driver to stop.

Make sense?

Thanks
Matt

I put the following

x=11
T=0.65
a=10.5 m/s(squared)
v=44 km/h

Now I would have to plug the above into one of these

v=V(not) + at

x=V(not)t + 1/2at(Squared)

V(Squared) =V(not) + 2ax


This is where I get confused.

I do not know which equation to use. I am thinking I would use
V(Squared) =V(not) + 2ax

Because it is the only one with an X

So if I plug in the variables I would get

44=0 + 2(10.5x)

44 = 21x

x=2

So the distance would be 2m?

Gosh I am confused lol

Thanks again !

 

[CFDFEAGURU]

You are solving for a distance, x, so use the equation that produces the distance, x, as the result.

Thanks
Matt

 

[Firebird]

You are solving for a distance, x, so use the equation that produces the distance, x, as the result.

Thanks
Matt

So that would be x= v(not)t+1/2at(squared)


?

 

[CFDFEAGURU]

Yes, that is the one. Also, as a check, make sure all of your units cancel to produce meters.

Thanks
Matt

 

[drizzle]

the driver's reaction time is not the time taken till the velocity reaches zero, you can find the time using the first equation:
v=v_o + at

[don’t forget to subtract the driver's reaction time from the calculated time]

 

[CFDFEAGURU]

Yes. That is correct.

Thanks
Matt