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2.Motion in Straight Line
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A particle starts from rest at $x=0\; m$ with an acceleration of $1 \,m / s ^2$. At $t = 5\;s _{ s }$ it receives an additional acceleration in the same direction as its motion. At $t =10\; s$ its speed and position are $v$ and $x$, respectively. Had the additional acceleration not been provided, its speed and position would have been $v _0$ and $x _0$, respectively. It is found that $x - x _0$ is $12.5 \,m$. Then one can conclude that $v - v _0$ is .............. $\,m / s$
A $5$
B$10$
C$15$
D$20$
(KVPY-2021)
Solution
(A)$x _{ B } \Rightarrow \frac{1}{2} \times(1) \times(5)^2 \Rightarrow 12 . .5 \,m$
$v _{ B } \Rightarrow 1 \times 5=5 \,m / s$
Let the additional acceleration provided at $t =$ 5 is $a$.
$\therefore$ new acceleration $= a +1$
$x \Rightarrow 12.5+\frac{1}{2}(a+1)(5)^2 \ldots(1)$
$v \Rightarrow 5+(a+1) 5 \ldots(2)$
if acceleration $=1$, then
$x_0 \Rightarrow 12.5+\frac{1}{2}(1)(5)^2 \ldots(3)$
$v_0 \Rightarrow 5+1 \times 5 \ldots(4)$
$x-x_0 \Rightarrow 12.5$ (given)
$\therefore 12.5 \Rightarrow \frac{1}{2}(a)(5)^2 \Rightarrow a=1$
Hence, $v – v _0 \Rightarrow a \times 5 \Rightarrow 5 \,m / s$
Standard 11
Physics
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