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(B) The height of the stone is given by the function $s(t) = 24t - 0.8t^2$. To find the velocity $v(t)$,we differentiate the position function with re

Vedclass · Accepted Answer

$1.6$

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(B) The height of the stone is given by the function $s(t) = 24t - 0.8t^2$. To find the velocity $v(t)$,we differentiate the position function with respect to time $t$: $v(t) = \frac{ds}{dt} = \frac{d}{dt}(24t - 0.8t^2) = 24 - 1.6t$. To find the acceleration $a(t)$,we differentiate the velocity function with respect to time $t$: $a(t) = \frac{dv}{dt} = \frac{d}{dt}(24 - 1.6t) = -1.6 \ m/s^2$. The negative sign indicates that the acceleration is directed downwards (opposing the upward motion). Therefore,the magnitude of the acceleration due to gravity on the surface of the moon is $1.6 \ m/s^2$.

$A$ stone,thrown vertically upward from the surface of the moon at a velocity of $24 \ m/s$,reaches a height of $s = 24t - 0.8t^2$ meters after $t$ seconds. The acceleration due to gravity in $m/s^2$ at the surface of the moon is:

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