A car decelerates from a speed of 64 km/hr to rest in 3.9 seconds. What is the magnitude of the acceleration in m/s^2 to one decimal place?

Question

A car decelerates from a speed of 64 km/hr to rest in 3.9 seconds.  What is the magnitude of the acceleration in m/s^2 to one decimal place?

abhisar · Answer

Hello @MoonlitFate !

moonlitfate · Answer

Hello @Abhisar . :D

abhisar · Answer

Do you know about equation of motions ?

moonlitfate · Answer

I don't think so?

abhisar · Answer

Is this equation familiar to you ?

$\sf s=ut+1/2at^2$

moonlitfate · Answer

No. :/

abhisar · Answer

aww..

moonlitfate · Answer

I'm guessing my lack of knowledge is just going to make this more difficult. :( But, I actually want to learn...

abhisar · Answer

It's ok everybody at some point lacks knowledge....and the good part is we can always learn ;)

moonlitfate · Answer

True.  I wish I would have studied over the summer instead of jumping in... to Physics and Calculus II.  SO difficult...

abhisar · Answer

So there are three equations of motion when the acceleration is constant,

$\sf v=u+at\s=ut+1/2at^2\v^2=u^2+2as$

abhisar · Answer

Here,
u=initial velocity,
v=final velocity,
a=accceleration,
s=distance covered and t=time taken

moonlitfate · Answer

Okay.  I'm just writing this down as you go.  Sill here.

abhisar · Answer

ok that's cool. For solving our question at the top, we are gonna use the second equation but let's first understand the question...u'll see how easy that question is !

abhisar · Answer

A car decelerates from a speed of 64 km/hr to rest in 3.9 seconds.  What is the magnitude of the acceleration in m/s^2 to one decimal place?

So, the initial velocity of the car is 64km/hr or 17.7 m/s. The decelerates finally to rest so the final velocity will be 0. Time taken for this is 3.9 seconds.

ok ?

moonlitfate · Answer

Yes, get going.  Copying this done, and re-reading to make sure I understand. :)

abhisar · Answer

ok, now we simply plug in the value into the first equation of motion.
v=0
u=17.7
t=3.9

=> 0=17.7 +a*3.9

Solving it for a, we get

-17.7/3.9=a = -4.38 m/s^2

$\therefore$ The acceleration is $\sf -4.38m/s^2$ or the deceleration is $\sf 4.38 m/s^2$

moonlitfate · Answer

Oh, that makes sense since. :)

abhisar · Answer

are you sure ?

moonlitfate · Answer

Yeah, can we do some more though? :)

abhisar · Answer

yeah sure ! and To know more about the equations, you can visit http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

moonlitfate · Answer

Thank you! :)  Can I type up the next one?

abhisar · Answer

Yeah sure !
Type the next one as a new post :)

moonlitfate · Answer

Alright.