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(C) The position of the particle is given by $x(t) = 10 + 6t - 3t^2$.To find the velocity,we differentiate the position with respect to time: $v(t) =

Vedclass · Accepted Answer

$27$

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(C) The position of the particle is given by $x(t) = 10 + 6t - 3t^2$.To find the velocity,we differentiate the position with respect to time: $v(t) = \frac{dx}{dt} = 6 - 6t$.Setting $v(t) = 0$,we find the particle stops at $t = 1 \ s$.Since the interval given is $t = 1 \ s$ to $t = 4 \ s$,the particle moves in only one direction (negative direction) after $t = 1 \ s$.At $t = 1 \ s$,the position is $x(1) = 10 + 6(1) - 3(1)^2 = 10 + 6 - 3 = 13 \ m$.At $t = 4 \ s$,the position is $x(4) = 10 + 6(4) - 3(4)^2 = 10 + 24 - 48 = -14 \ m$.The distance travelled is the magnitude of the displacement: $d = |x(4) - x(1)| = |-14 - 13| = |-27| = 27 \ m$.

The position of a particle moving along the $x$ axis is given by the equation $x = (10 + 6t - 3t^2) \ m$. The distance travelled by the particle from $t = 1 \ s$ to $t = 4 \ s$ is: (in $m$)

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