(C) The displacement equation is given by $s = 13.8t - 4.9t^2$.The velocity $v$ is the rate of change of displacement with respect to time,given by $v

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(C) The displacement equation is given by $s = 13.8t - 4.9t^2$.The velocity $v$ is the rate of change of displacement with respect to time,given by $v = \frac{ds}{dt}$.Differentiating $s$ with respect to $t$:$v = \frac{d}{dt}(13.8t - 4.9t^2) = 13.8 - 9.8t$.To find the velocity at $t = 1$ second,substitute $t = 1$ into the velocity equation:$v(1) = 13.8 - 9.8(1) = 13.8 - 9.8 = 4.0 \text{ } m/s$.Thus,the velocity at $t = 1$ second is $4 \text{ } m/s$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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$A$ stone thrown vertically upwards has its motion described by the equation $s = 13.8t - 4.9t^2$,where $s$ is in meters and $t$ is in seconds. The velocity at $t = 1$ second will be ...... $m/s$.

A
$3$
B
$5$
C
$4$
D
None of these.

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$A$ stone thrown vertically upwards has its motion described by the equation $s = 13.8t - 4.9t^2$,where $s$ is in meters and $t$ is in seconds. The velocity at $t = 1$ second will be ...... $m/s$.

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