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A particle starts from origin at time $t=0$ and moves in positive $x$-direction. Its velocity $\mathbf{v}$ varies with times as $\mathbf{v}=10 \hat{t} \mathrm{~cm} / \mathrm{s}$. The distance covered by the particle in $8 \mathrm{~s}$ will be
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Verified Answer
The correct answer is:
$320 \mathrm{~cm}$
Given that, $t=0, \mathbf{v}=10 \hat{i} \mathrm{~cm} / \mathrm{s}$
So, acceleration, $a=\frac{d v}{d t}=10 \hat{\mathbf{i}} \mathrm{cm} / \mathrm{s}$
According to second equation of motion,
$$
\begin{array}{rlrl}
& & s & =u t+\frac{1}{2} a t^2 \\
& u & u & =0 \\
& s & =1 / 2 \times 10 \times(8)^2=320 \\
& s & =320 \mathrm{~cm}
\end{array}
$$
So, acceleration, $a=\frac{d v}{d t}=10 \hat{\mathbf{i}} \mathrm{cm} / \mathrm{s}$
According to second equation of motion,
$$
\begin{array}{rlrl}
& & s & =u t+\frac{1}{2} a t^2 \\
& u & u & =0 \\
& s & =1 / 2 \times 10 \times(8)^2=320 \\
& s & =320 \mathrm{~cm}
\end{array}
$$
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