Question

the equation for free fall at the surface of some planet (s in meters, t in seconds) is s = 10.01t². how long does it take a rock falling from rest to reach a velocity of 25.4 m/s on this planet? it takes 1.3 sec for a rock falling from rest to reach a velocity of 25.4 m/s. (round to the nearest tenth as needed.)

Explanation:

Step1: Recall the velocity - displacement relationship

Velocity $v$ is the derivative of displacement $s$ with respect to time $t$. Given $s = 10.01t^{2}$, by the power - rule of differentiation $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v=\frac{ds}{dt}=2\times10.01t = 20.02t$.

Step2: Solve for time $t$

We know that $v = 25.4$ m/s. Substitute $v$ into the velocity formula $v = 20.02t$. Then $t=\frac{v}{20.02}$.
Substitute $v = 25.4$ into the formula: $t=\frac{25.4}{20.02}\approx1.3$ s.

Answer:

$1.3$ s