logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
0 votes

A bus starts from rest with an acceleration of $ 1\;ms^{-2}$ A man who is $48\;m$ behind the bus starts with a uniform velocity of $ 10\;ms^{-1}$. The minimum time after which the man will catch the bus is

$\begin{array}{1 1} 4.8\;s \\ 8\;s \\ 10\;s \\12\;s \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Answer : $ 8\;s$
$10t =48+\large\frac{1}{2} $$+ 1 \times t^2 $
or $ t^2-20t+96=0$
or $t^2-8t-12t+96=0$
or $t(t-8) -12 (t-8) -12 (t-8)=0$
or $(t-12) (t-8) =0$
or $t=8s$ or $12\;s$
But we are interested in minimum time.
answered Aug 19, 2014 by meena.p
Your answer is correct, no doubt in it but write the exact statement  before solving any problem.  You haven't explained  the frame of reference  between man and the bus.
Good Luck.
 

Related questions

Ask Question
Tag:MathPhyChemBioOther
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...