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A body starts from rest and moves with uniform acceleration. If the distance travelled by it in the first $2 \mathrm{~s}$ is $x_1$ and in the next $2 \mathrm{~s}$ is $x_2$, then $x_1$ and $x_2$ are related as
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Verified Answer
The correct answer is:
$3 x_1=x_2$
If $a$ be the acceleration of the body, then distance travelled by the body in $2 \mathrm{~s}$ is given as,
$$
\begin{gathered}
x_1=u t+\frac{1}{2} a t^2=0 \times 2+\frac{1}{2} a \times(2)^2 \\
x_1=2 a
\end{gathered}
$$
Velocity of body at the end of $2 \mathrm{~s}$ is given as,
$$
\begin{aligned}
& v=u+a t=0+a \times 2 \\
& v=2 a
\end{aligned}
$$
Distance travelled by the body in next $2 \mathrm{~s}$ is given as,
$$
\begin{aligned}
& x_2=v t+\frac{1}{2} a t^2 \\
& =2 a \times 2+\frac{1}{2} a \times(2)^2 \text { [from Eq. (ii)] } \\
& =4 a+2 a \\
& x_2=6 a=3 \times 2 a \\
& \Rightarrow \quad x_2=3 \times x_1 \quad \text { [from Eq. (i)] } \\
& \Rightarrow \quad x_2=3 x_1 \\
&
\end{aligned}
$$
$$
\begin{gathered}
x_1=u t+\frac{1}{2} a t^2=0 \times 2+\frac{1}{2} a \times(2)^2 \\
x_1=2 a
\end{gathered}
$$
Velocity of body at the end of $2 \mathrm{~s}$ is given as,
$$
\begin{aligned}
& v=u+a t=0+a \times 2 \\
& v=2 a
\end{aligned}
$$
Distance travelled by the body in next $2 \mathrm{~s}$ is given as,
$$
\begin{aligned}
& x_2=v t+\frac{1}{2} a t^2 \\
& =2 a \times 2+\frac{1}{2} a \times(2)^2 \text { [from Eq. (ii)] } \\
& =4 a+2 a \\
& x_2=6 a=3 \times 2 a \\
& \Rightarrow \quad x_2=3 \times x_1 \quad \text { [from Eq. (i)] } \\
& \Rightarrow \quad x_2=3 x_1 \\
&
\end{aligned}
$$
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