How long to the nearest second does it take a car to accelerate from rest at 3 m/s^2 to cover a distance of 119 meters?

Question

abhisar · Answer

u=0
s=119
a=3

Plug in the values into $\sf s=ut+1/2at^2$ and find the value of t

moonlitfate · Answer

@Abhisar -- Hmm, I don't think I got the right thing.

moonlitfate · Answer

Wow, defintely didn't get that..

abhisar · Answer

u got how i did the above calculation ?

moonlitfate · Answer

You have to isolate t.  Then again my math skills could just be really rusty..

abhisar · Answer

ok i am elaborating it...it's easy

$\sf \because~s=ut+1/2at^2\ \Rightarrow 119=0\times t+0.5\times 3\times t^2\ 119=1.5\times t^2\ t^2=119/1.5\t^2=79.33\t=\sqrt{79.33}=8.9s$

abhisar · Answer

My first answer was wrong !

moonlitfate · Answer

Ooooh, I was putting the at^2 in the denominator of the fraction.

abhisar · Answer

Is it clear now ?

moonlitfate · Answer

I got 8.906, and since it's the nearest second, it would be 9?

abhisar · Answer

yes

moonlitfate · Answer

:)  That's so much easier than what my book has.

abhisar · Answer

:)

moonlitfate · Answer

I'll be posting more questions.  I need th epractice, lol.

abhisar · Answer

ok :)