Answer
a. $24-1.6t$ m/s and $-1.6$ m/s$^2$
b. $15$ s
c. $180$ m
d. $4.4$ s and $25.6$ s
e. $30$ s.
Work Step by Step
a. Given $s(t)=24t-0.8t^2$, we can find the velocity as $v(t)=s'(t)=24-1.6t$ m/s and acceleration $a(t)=v'(t)=-1.6$ m/s$^2$.
b. At the highest point $v(t)=0$, which gives $24-1.6t=0$ or $t=15$ s.
c. At $t=15$, we have $s(15)=24(15)-0.8(15)^2=180$ m.
d. Letting $s(t)=180/2=90$, we have $24t-0.8t^2=90$, which gives $t^2-30t+112.5=0$. Solving this quadratic equation numerically or graphically, we can obtain $t\approx4.4$ sec and $t\approx25.6$ sec. There are two time values because the object will reach this height both going up and falling down.
e. We need to calculate the total time the rock spends before it hits the ground. Letting $s(t)=0$, we have $24t-0.8t^2=0$ which gives $t=0$ and $t=30$ sec. As $t=0$ means the starting point, we conclude that the rock will be aloft for $30$ seconds.
