A particle moves in a straight line with a constant acceleration. It changes its velocity from $$10\,m{s^{ - 1}}$$ to $$20\,m{s^{ - 1}}$$ while passing through a distance $$135\,m$$ in $$t$$ $$\sec.$$ The value of $$t$$ is
A.
10
B.
1.8
C.
12
D.
9
Answer :
9
Solution :
Using $${v^2} - {u^2} = 2\,as$$
$${\left( {20} \right)^2} - {\left( {10} \right)^2} = 2 \times a \times 135 \Rightarrow \frac{{300}}{{270}} = a = \frac{{10}}{9}$$
Now, using $$v - u = at$$
$$\eqalign{
& 20 - 10 = \frac{{10}}{9} \times t \cr
& \Rightarrow t = 9\,s \cr} $$
Releted MCQ Question on Basic Physics >> Kinematics
Releted Question 1
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