The position of a particle moving along the x | vedclass

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

Vedclass · Accepted Answer

$2.5$

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(B) The position of the particle is given by $x = 10t - 2t^2$.To find the velocity $(v)$,we differentiate the position with respect to time $(t)$:$v = \frac{dx}{dt} = \frac{d}{dt}(10t - 2t^2) = 10 - 4t$.For the particle to momentarily come to rest,its velocity must be zero:$v = 0$.Substituting the expression for velocity:$10 - 4t = 0$.$4t = 10$.$t = \frac{10}{4} = 2.5 \, s$.Thus,the particle comes to rest at $t = 2.5 \, s$.

The position of a particle moving along the $x$-axis is given by $x = 10t - 2t^2$. Then the time $(t)$ at which it will momentarily come to rest is .......... $s$.

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