If s=60 t-5 t^2 denotes the distance covered | vedclass

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(D) Given the distance function $s = 60t - 5t^2$.The velocity $v$ is the rate of change of distance with respect to time $t$,given by $v = \frac{ds}{d

Vedclass · Accepted Answer

$180$

Vedclass · Answer

(D) Given the distance function $s = 60t - 5t^2$.The velocity $v$ is the rate of change of distance with respect to time $t$,given by $v = \frac{ds}{dt}$.$\frac{ds}{dt} = \frac{d}{dt}(60t - 5t^2) = 60 - 10t$.The particle comes to rest when its velocity $v = 0$.Setting $60 - 10t = 0$,we get $10t = 60$,which implies $t = 6 	ext{ s}$.Now,we calculate the distance covered at $t = 6 	ext{ s}$:$s(6) = 60(6) - 5(6)^2$$s(6) = 360 - 5(36)$$s(6) = 360 - 180 = 180 	ext{ units}$.Thus,the distance covered before coming to rest is $180 	ext{ units}$.

If $s=60 t-5 t^2$ denotes the distance covered by a particle in time $t$,then the distance it covers before coming to rest is.... units.

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