The position of a body moving along the x-axi | vedclass

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(A) Given position: $x(t) = t^2 - 4t + 6$.Velocity $v(t) = \frac{dx}{dt} = 2t - 4$.The body comes to rest when $v(t) = 0$,which gives $2t - 4 = 0$,so

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$5$

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(A) Given position: $x(t) = t^2 - 4t + 6$.Velocity $v(t) = \frac{dx}{dt} = 2t - 4$.The body comes to rest when $v(t) = 0$,which gives $2t - 4 = 0$,so $t = 2 \, s$.Since the motion reverses at $t = 2 \, s$,we calculate the distance in two parts: from $t = 0$ to $t = 2 \, s$ and from $t = 2 \, s$ to $t = 3 \, s$.At $t = 0 \, s$: $x(0) = 0^2 - 4(0) + 6 = 6 \, m$.At $t = 2 \, s$: $x(2) = 2^2 - 4(2) + 6 = 4 - 8 + 6 = 2 \, m$.At $t = 3 \, s$: $x(3) = 3^2 - 4(3) + 6 = 9 - 12 + 6 = 3 \, m$.Distance from $t = 0$ to $t = 2 \, s$ is $|x(2) - x(0)| = |2 - 6| = 4 \, m$.Distance from $t = 2$ to $t = 3 \, s$ is $|x(3) - x(2)| = |3 - 2| = 1 \, m$.Total distance = $4 \, m + 1 \, m = 5 \, m$.

The position of a body moving along the $x$-axis at time $t$ is given by $x = (t^2 - 4t + 6) \, m$. The distance travelled by the body in the time interval $t = 0$ to $t = 3 \, s$ is ...... $m$.

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